Matrices

MathsWorks for Teachers

MathsWorks for Teachers Series editor David Leigh-Lancaster

This book provides mathematics teachers with an elementary introduction to matrix algebra and its uses in formulating and solving practical problems, solving systems of linear equations, representing combinations of affine (including linear) transformations of the plane and modelling finite state Markov chains. The basic theory in each of these areas is explained and illustrated using a broad range of examples. A feature of the book is the complementary use of technology, particularly computer algebra systems, to do the calculations involving matrices required for the applications. A selection of student activities with solutions and text and web references are included throughout the book.

Pam Norton

Matrices are used in many areas of mathematics, and have also have applications applications in diverse in diverse areas such areas as such engineering, as engineering, computer computer graphics,graphics, image processing, image processing, physicalphysical sciences, sciences, biological biological sciences and sciences socialand sciences. social Powerful sciences. Powerful calculators calculators and computers and computers can now can now carry outcarry complicated out complicated and difficult and numeric difficult numeric and algebraic and algebraic computations computations involving involving matrix methods, matrix and methods, such technology and such technology is a vital tool is ain vital related tool real-life, in relatedproblemreal-life, problem-solving solving applications. applications.

MATRICES

Matrices Pam Norton

Series overview

MathsWorks for Teachers has been developed to provide a coherent and contemporary framework for conceptualising and implementing aspects of middle and senior mathematics curricula.

Matrices

Titles in the series are: Functional Equations David Leigh-Lancaster Contemporary Calculus Michael Evans Matrices Pam Norton Foundation Numeracy in Context David Tout & Gary Motteram Data Analysis Applications Kay Lipson Complex Numbers and Vectors Les Evans

Functional Equations

Contemporary Calculus

David Leigh-Lancaster

Michael Evans

MathsWorks for Teachers

MathsWorks for Teachers

Matrices Pam Norton

MathsWorks for Teachers

Foundation Numeracy in Context

Data Analysis Applications

Complex Numbers and Vectors

David Tout and Gary Motteram

Kay Lipson

Les Evans

MathsWorks for Teachers

MathsWorks for Teachers

MathsWorks for Teachers

Pam Norton

ISBN 978-0-86431-508-3

9

780864 315083

MathsWorks for Teachers

Matrices Pam Norton

MathsWorks for Teachers

First published 2007 by ACER Press Australian Council for Educational Research Ltd 19 Prospect Hill Road, Camberwell, Victoria, 3124 Copyright © 2007 Pam Norton and David Leigh-Lancaster All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, no part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the written permission of the publishers. Edited by Marta Veroni Cover design by FOUNDRY Typography, Design & Visual Dialogue Text design by Robert Klinkhamer Typeset by Desktop Concepts P/L, Melbourne Printed by Shannon Books Australia Pty Ltd National Library of Australia Cataloguing-in-Publication data: Norton, Pam. Matrices. For teachers of senior secondary mathematics courses. ISBN 9780864315083. 1. Matrices – Study and teaching (Secondary). I. Leigh-Lancaster, David. II. Australian Council for Educational Research. III. Title. (Series: Mathsworks for teachers). 512.94340712 Visit our website: www.acerpress.com.au

Contents Introduction  v About the author  vi

1

An introduction to matrices  1 History  2 Matrices in the senior secondary mathematics curriculum  5

2

Rectangular arrays, matrices and operations  11 Definition of a matrix  15 Operations on matrices  16 Addition and subtraction of two matrices  17 Multiplication by a number (scalar multiple)  17 Structure properties of matrix addition and scalar multiplication  19 Matrix multiplication  20 Zero and identity matrices  23 The transpose of a matrix  25 The inverse of a matrix  27 Applications of matrices  30

3

Solving systems of simultaneous linear equations  37 Solving systems of simultaneous linear equations using matrix inverse  43 The method of Gaussian elimination  50 Systems of simultaneous linear equations in various contexts  59

Contents

4

Transformations of the cartesian plane  75 Linear transformations  76 Linear transformation of a straight line  81 Linear transformation of a curve  87 Standard types of linear transformations  89 Composition of linear transformations  101 Affine transformations  103 Composition of affine transformations  104

5

Transition matrices  111 Conditional probability  111 Transition probabilities  112 The steady-state vector  119 Applications of transition matrices  121

6

Curriculum connections  136

7

Solution notes to student activities  141



References and further reading  163 Notes  165

Introduction MathsWorks is a series of teacher texts covering various areas of study and topics relevant to senior secondary mathematics courses. The series has been specifically developed for teachers to cover helpful mathematical background, and is written in an informal discussion style. The series consists of six titles: • Functional Equations • Contemporary Calculus • Matrices • Data Analysis Applications • Foundation Numeracy in Context • Complex Numbers and Vectors Each text includes historical and background material; discussion of key concepts, skills and processes; commentary on teaching and learning approaches; comprehensive illustrative examples with related tables, graphs and diagrams throughout; references for each chapter (text and web-based); student activities and sample solution notes; and a bibliography. The use of technology is incorporated as applicable in each text, and a general curriculum link between chapters of each text and Australian state and territory as well as selected overseas courses is provided. A Notes section has been provided at the end of the text for teachers to include their own comments, annotations and observations. It could also be used to record additional resources, references and websites.



A b o u t t h e a u t h o r Pam Norton is an experienced lecturer in university level mathematics. Her mathematical interests are in the applications of mathematics to sport, and her educational interests are in the use of technology in the teaching and learning of mathematics. She has been involved in the setting and assessing of examinations and extended assessment tasks in mathematics and curriculum review at both secondary and tertiary levels.

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A n i n t r o d u c t i o n t o m a t ric e s Throughout history people have collected and recorded various data using sets (unordered lists), vectors (ordered one-dimensional lists) and matrices (ordered two-dimensional lists of lists, tables or rectangular arrays). Today arrays and tables of numbers and other information are found widely in everyday life. For example, sports ladders give numbers of wins, losses, draws, points and other information for teams in a competition. Each day stock tables are given in local newspapers. Results of opinion polls are usually given in table form in newspapers. Information of many forms is held in tables, and the spreadsheet is an electronic digital technology that is now widely used in everyday life wherever data is to be entered, stored and manipulated using tables and rectangular arrays. The word matrix comes from the Latin word for ‘womb’. The term matrix is also used in areas other than mathematics, and generally means an environment, milieu, substance or place in which something is cast, shaped or developed. It has become conventional in contemporary mathematics to describe a matrix in terms of the specification of its elements by reference to row by column, following the use of this form in developments by 19th century mathematicians, but it is interesting to speculate whether an element of a matrix would most likely have been referenced by column first then by row, had computerised spreadsheets been around prior to the modern development of matrices! Indeed, Chinese mathematicians who first recorded the use of matrices used an early form of computer algebra, the counting board, and did reference elements by column before row (see Hoe, 1980).



MathsWorks for Teachers Matrices

Hi s t o r y In the history of mathematics, the explicit and formal development of matrices is a relatively modern invention. Katz (1993), Grattan-Guinness (1994) and Fraleigh and Beauregard (1995) provide useful historical background, as do the websites referenced at the end of this chapter. Although it is evident that ancient civilisations such as the Babylonians and Chinese were able to solve systems of simultaneous linear equations, it is not clear if systems with multiple solutions or no solutions were considered by them. The Egyptians were able to solve simple linear equations directly and by guessing an answer and adjusting it to the correct one. The origins of matrices are found in the study of systems of simultaneous linear equations, and can be traced back to the Chinese in the Han Dynasty about 200 BCE. The Han dynasty, established in about 210 BCE, developed two important texts for mathematical education: Zhoubi suanjing (Arithmetical Classic of the Gnomon and the Circular Paths of Heaven) and Jiuzhang suanshu (Nine Chapters on the Mathematical Art). In the eighth chapter of the latter, systems of simultaneous linear equations were solved with the aid of a counting board by arranging the equations in tabular form, with each column containing the coefficients and the constant term for one of the equations. The solution was then obtained by multiplying and subtracting columns to get a triangular form, followed by back-substitution. English translations and commentary on these texts have recently become available (see Kangshen, Crossley & Lun, 1999). Systems of simultaneous linear equations arise in many areas—economics, social sciences, medicine, engineering, biological and physical sciences and mathematics. The most useful method for solving such systems is known as Gaussian elimination. It was first used in ‘modern’ times by Gauss in 1809 to solve six simultaneous equations in six unknowns which he built while studying the orbit of the asteroid Pallas. Gauss simply developed the method first documented by the Chinese in about 200 BCE. Not only does this method have historical significance, but it is also the basis for the best direct methods for programming a computer to solve such systems today. While software programs such as MatLab have been specifically developed for highspeed numerical computation with high-order matrices, other programs and technologies such as spreadsheets, computer algebra systems (CAS) and graphics and CAS calculators can also carry out matrix computations with matrices that are too large for efficient by hand computation. CAS technologies can also carry out computations with algebraic as well as



chapter 1 An introduction to matrices

numerical elements for matrices. Where quick and accurate computations are required, modern calculator and computer technologies are indispensable tools. Gauss’s method was extended to become known as the Gauss–Jordan method for solving systems of simultaneous linear equations. Jordan’s contribution to this method is the incorporation of a systematic technique for back-substitution. The Gauss–Jordan method is used to obtain the reduced row echelon form of a matrix and hence to solve a system of simultaneous linear equations directly. The Jordan part was first described by Wilhelm Jordan, a German professor of geodesy, in the third edition (1888) of his Handbook of Geodesy. He used the method to solve symmetric systems of simultaneous linear equations arising out of a least squares application in geodesy. It is perhaps somewhat surprising that the idea of a matrix did not evolve until well after that of a determinant. In 1545, Cardan gave a rule for solving a system of two linear equations, which is essentially Cramer’s rule using determinants for solving a 2 × 2 system. The idea of a determinant appeared in Japan and Europe at approximately the same time. Seki Kowa in a manuscript in 1683 introduced the notion of a determinant (without using this name for them). He described their use and showed how to find determinants of matrices up to order 5 × 5. In the same year in Europe, Leibniz also introduced the idea of a determinant to explain when a system of equations had a solution. In 1750, Gabriel Cramer published his Introduction to the Analysis of Algebraic Curves. He was interested in the problem of determining the equation of a plane curve of a given degree that passes through a certain number of points. He stated a general rule, now known as Cramer’s rule, in an appendix, but did not explain how it worked. The term ‘determinant’ was first introduced by Gauss in 1801 while discussing quadratic forms. Binary quadratic forms are expressions such as ax 2 + 2bxy + cy2, where x and y are variables and a, b and c coefficients, which can be represented in matrix form as:

6 x y@ =

a b x G= G b c y

Gauss considered linear substitutions for the variables x and y of the form x = ax1 + by1 y = gx1 + hy1

and composition of substitutions, which led to matrix multiplication.



MathsWorks for Teachers Matrices

Augustin-Louis Cauchy, in 1812, published work on the theory of determinants, both using the term determinant and the abbreviation (a1,n) to stand for the symmetric system a1, 1 a1, 2 f a1, n a2, 1 a2, 2 f a2, n



h h h h an, 1 an, 2 f an, n

that is associated with the determinant. Although many of the basic results in calculating determinants were already known, he introduced work on minors and adjoints, and the procedure for calculating a determinant by expanding along any row or down any column (now called the Laplace expansion). It was not until 1850 that James Joseph Sylvester used the term matrix to refer to a rectangular array of numbers. He spent most of his time studying determinants that arose from matrices. Soon after, Arthur Cayley showed that matrices were useful to represent systems of simultaneous linear equations, with the notion of an inverse matrix for their solution. In 1858 Cayley gave the first abstract definition of a matrix, and subsequently developed the algebra of matrices, defining the operations of addition, multiplication, scalar multiplication and inverse. He also showed that every matrix satisfies its characteristic equation, a result which is now known as the Cayley–Hamilton theorem. Cayley proved this theorem for 2 × 2 matrices, and had checked the result for 3 × 3 matrices, while William Rowan Hamilton had proved the special case for 4 × 4 matrices with his investigations into the quaternions. Georg Frobenius proved it for the general case in 1878. Many other mathematicians have contributed to the theory of matrices and determinants, including Etienne Bezout, Colin Maclaurin, AlexandreTheophile Vandermonde, Pierre Simon Laplace, Joseph Louis Lagrange, Ferdinand Gotthold Eisenstein, Camille Jordan, Jaques Sturm, Karl Gustav Jacob Jacobi, Leopold Kronecker and Karl Weierstrass. Markov chains are named after the Russian mathematician Andrei Andreevich Markov, who first defined them in 1906 in a paper dealing with the Law of Large Numbers. He used examples from literary books, with the two possible states being vowels and consonants. To illustrate his results, he did a statistical study of the alternation of vowels and consonants in Pushkin’s Eugene Onegin. Matrices and matrix methods are now used in many areas of practical and theoretical application. The Harvard economist Wassily W. Leontief was



chapter 1 An introduction to matrices

awarded the Nobel Prize in Economics for his work on input–output models, which relied heavily on matrices and solving systems of simultaneous linear equations. Matrices are also used extensively in business optimisation contexts, in particular where networks (graph theory) are applied to problems involving representation, connectedness and allocation. The mathematician Olga Taussky Todd developed matrix applications to analyse vibrations on airplanes during World War II, and made an important contribution to the development and application of matrix theory. The development of the electronic digital computer has had a big impact on the use of matrices in many areas. Matrix methods are used extensively in computer graphics, a developing area especially driven by the demands of the movie and computer games industries. Matrix methods are also used extensively in the communication industry (especially for encryption and decryption), in engineering and the sciences, and in economic modelling and industry. In mathematics, the study of matrices and determinants is part of linear algebra, and it is recognised that matrices and their natural operations provide models for the algebraic structures of a vector space and of a noncommutative ring.

Ma t ric e s i n t h e s e n i o r s e c o n d ar y m a t h e m a t ic s c u rric u l u m During the 1960s and 1970s matrices began to be incorporated into aspects of the senior secondary mathematics curriculum, in particular following the innovations of the new mathematics program from the 1950s (see, for example, Allendoerfer & Oakley, 1955; Adler, 1958), and the increasing use of them as a tool for business-related applications. In curriculum terms this can also be related to a greater emphasis on discrete mathematics in curriculum design during the second half of the 20th century. There have been several purposes for which matrices have been introduced into the senior secondary mathematics curriculum: • to represent and solve systems of m simultaneous linear equations in n variables, where m, n ≥ 2 • to represent and apply transformations, and combinations of transformations of the cartesian plane, in particular considering those subsets of the cartesian plane that represent graphs of functions and other relations



MathsWorks for Teachers Matrices

• as arrays to model and manipulate data related to practical situations such as stock inventories, sales and prices • to represent and compute states and transitions between states, for example population modelling, tax scales and conditional probabilities, including Markov chains • as a mathematical tool for analysis in graph theory (networks) and game theory • to carry out numerical computation such as for approximation of irrational real numbers • to provide a model for abstract mathematical structures such as noncommutative rings and vector spaces • to provide a model for other mathematical entities such as complex numbers and vectors The incorporation of matrices within the school mathematics curriculum has been both as objects for investigation in their own right, and for their instrumental application in particular contexts. For example, in The New Mathematics (Adler, 1958) matrices are introduced for analysis of linear transformations of the plane, then considered as an algebra in their own right, and finally used as a model for the complex numbers. In The Principles of Mathematics (Allendoerfer & Oakley, 1955) matrices are briefly introduced as an example of a non-commutative ring. During the 1970s in Aggregating Algebra (Holton & Pye, 1976), matrices are introduced in the context of systems of three simultaneous linear equations in three unknowns with an emphasis on geometric interpretation of the solutions. Similarly, in School Mathematics Research Foundation: Pure Mathematics (Cameron et al., 1970), matrices are introduced in the context of linear transformations of the plane and then applied to the solution of systems of simultaneous linear equations (with a detailed discussion on elementary matrices) and finally a brief consideration of matrix algebras. Around this time matrices also began to be used increasingly for practical applications within discrete business-oriented mathematics courses (see, for example, Kemeny et al., 1972) including applications of transition matrices. During the 1970s and early 1980s, within pure mathematics-oriented senior secondary courses, especially those intended for students with interest and aptitude in higher level mathematics, such as Pure Mathematics (Fitzpatrick & Galbraith, 1971), the use of matrices covered transformations, solution of systems of simultaneous linear equations and investigation of mathematical structures (groups of transformations of the plane, complex numbers and rings). In some cases this also included consideration of determinants for



chapter 1 An introduction to matrices

equation solving (Cramer’s rule) and finding inverses. However, a comprehensive study of determinants is a substantial area of mathematical investigation of its own, and it has typically not been included in senior secondary mathematics curricula in its own right, but rather in relation to the study of matrices (see, for example, Hodgson & Patterson, 1990). In terms of contemporary senior secondary curricula, matrices are covered in, for example, Further Mathematics and Mathematical Methods (CAS) studies in Victoria, Australia, and in the Further Pure AS module: Matrices and Transformations in the United Kingdom. Matrix computations had the advantage, within the contexts typically used in senior secondary mathematics courses, of requiring relatively simple arithmetic calculations. However, for matrices of order 3 × 3 and greater, the extent of these calculations meant that reliability of correct calculation became problematic, and much of the time working on matrices was spent on learning algorithms for the necessary computations and then carrying these out mentally, by hand, or possibly with the assistance of arithmetic or scientific calculators. A consequence of this was that only matrices of small orders were used, and computations involving finding inverses tended to be associated with more formal rather than practically oriented senior secondary mathematics courses. The advent from the late 1980s of powerful and readily accessible mathematically able technology, initially in the form of mini-computers and later as desk-top computers with software such as spreadsheets, numerical computation software and CAS, and subsequently hand-held graphics and CAS calculators, has meant that the computational load associated with matrix work can be carried out by these technologies. Thus, without the time and reliability constraints imposed by access only to mental and/or by hand arithmetic calculation, senior secondary mathematics students across a broad range of courses can utilise matrices, including those of higher orders, for a variety of purposes. The efficient and effective use of these technologies requires students to have a sound conceptual understanding of key matrix definitions (such as order, row, column, types of matrices) and conditions for computations (such as conformability, existence of inverses), as well as practical mental and by hand facility with computation in simple cases so that they can understand what computations are taking place, verify the reasonableness of results, and anticipate the nature of these results to verify their mathematical working. While matrices continue to have a strong role in providing a unifying abstract structure in contemporary senior secondary mathematics curricula, their



MathsWorks for Teachers Matrices

instrumental value in practical modelling and applications can be enhanced in conjunction with the use of modern technology (see, for example, Kissane 1997; Garner, McNamara & Moya, 2004). Although purpose-designed computational software such as Matlab is used in complicated real-life applications where high speed computations involve matrices of very high orders, general CAS such as Derive, Maple and Mathematica and spreadsheets also have powerful matrix functionalities. Student versions of these general CAS, or graphics and CAS calculators can be used by students for examples suitable for senior secondary mathematics courses. The author has used the CAS software Derive (which also underpins computation in the TI-89, TI-92 and Voyage 200 series of hand-held CAS calculators) for matrix computations in this text. More recent hand-held technology, such as the CASIO Classpad 300 series, the TI-nspire CAS+ and the HP 50G, have very user-friendly template-based matrix functionality. SUMM A R Y

• Various data can be collected and recorded using sets (unordered lists), vectors (ordered one-dimensional lists) and matrices (ordered two-dimensional lists of lists). • Matrices are ordered rectangular arrays, or lists of equal-sized lists, that are constructed by arranging data in rows and columns. • Rectangular tables are simple examples of matrices. • The origins of matrices can be found in Chinese mathematical instructional texts from around 200 BCE, with application to the solution of systems of simultaneous linear equations. • In Europe, matrices arose from Gauss’s work in the early 19th century on solving systems of simultaneous linear equations. This work provides a general method (known as Gaussian elimination). The term matrix was first applied to a rectangular array of numbers by Sylvester in 1850. • Gaussian elimination, and related methods, form the basis for algorithms used by modern technology to solve systems of simultaneous linear equations. • The idea of a determinant arose independently in both Japan and Europe in the latter part of the 17th century. • The modern theory of matrices and determinants was developed significantly in the latter part of the 19th century and the early 20th century.



chapter 1 An introduction to matrices

SUMM A R Y (Cont.)

• Matrices began to be incorporated in senior secondary mathematics curricula from the 1960s, as discrete mathematics began to have a more significant role in curriculum design • Matrices have been used in the senior secondary mathematics curriculum to: – solve simple systems of simultaneous linear equations – apply transformations of the plane to sets of points – solve practical problems in which information can be modelled and manipulated using matrices and matrix operations – analyse transition states, such as in simple Markov chains – provide an example of an abstract mathematical structure – model complex numbers and vectors. • From the 1960s through to the early 1980s, calculation of matrix operations in the senior secondary mathematics was generally carried out by hand (possibly with the assistance of a scientific calculator) and hence was restricted to applications involving matrices of low order. • Access to spreadsheets, numeric processors, graphics and CAS calculators, and CAS software as enabling technology in the senior secondary mathematics curriculum from the late 1980s has permitted a broader range of applications involving higher order matrices to be addressed.

References Adler, I 1958, The new mathematics, John Day, New York. Allendoerfer, CB & Oakley, CO 1955, Principles of mathematics, McGraw-Hill, New York. Cameron, N, Clements, K, Green, LJ & Smith, GC 1970, School Mathematics Research Foundation: Pure mathematics, Chesire, Melbourne. Fitzpatrick, JB & Galbraith, P 1971, Pure mathematics, Jacaranda, Milton, Queensland. Fraleigh, JB & Beauregard, AR 1995, Linear algebra (3rd edition with historical notes by VJ Katz), Addison-Wesley, Reading, MA. Garner, S, McNamara, A & Moya, F 2004, CAS analysis supplement for Mathematical Methods CAS Units 3 and 4, Pearson, Melbourne. Grattan-Guinness, I (ed.) 1994, Companion encyclopedia of the history and philosophy of the mathematical sciences, volume 1, Routledge, London.



MathsWorks for Teachers Matrices Hodgson, B & Patterson, J 1990, Change and approximation, Jacaranda, Melbourne. Hoe, J 1980, ‘Zhu Shijie and his jade mirror of the four unknowns’, in JN Crossley (ed.), First Australian conference on the history of mathematics: Proceedings, Monash University, Clayton. Holton, DA & Pye, W 1976, Aggregating algebra, Holt, Rhinehart and Winston, Sydney. Kangshen, S, Crossley, JN & Lun, A 1999, The nine chapters on the mathematical art: Companion and commentary, Oxford University Press, Oxford. Katz, VJ 1993, A history of mathematics—an introduction, Harper-Collins, Reading, MA. Kemeny, JG, Scheifer, A, Snell, JL & Thompson, GC 1972, Finite mathematics with business applications, Prentice-Hall, Englewood Cliffs, NJ. Kissane, B 1997, More mathematics with a graphics calculator, Mathematical Association of Western Australia, Claremont.

Websites http://www.ualr.edu/lasmoller/matrices.html – History Department, University of Arkansas at Little Rock This website includes a concise overview of the history of matrices with a good list of links to related sites, including online applications for matrix computation. http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Matrices_and_ determinants.html – School of Mathematics and Statistics, University of St Andrews Scotland This website contains a comprehensive historical coverage of matrices, determinants and related mathematics from ancient Babylonia and China through to the modern era, with extensive cross references to key mathematicians and related topics.

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R e c t a n g u l ar arra y s , m a t ric e s a n d o p e ra t i o n s The purpose of this chapter is to provide some practical contexts to motivate the definition of matrices and their features, to introduce several operations on matrices, and to discuss their properties. When a new mathematical structure is introduced, students often need to explore the rationale behind this process, and find a concrete model for the elements of the structure and related operations defined on these elements that they can interpret in context. They can then subsequently explore certain generalisations or specialisations related to this structure as they become more confident with its elements and operations in their own right. There are many different ways in which arrays can be made in twodimensions, typically based on regular geometric shapes such as polygons or circles. Such arrays can be devised by placing a finite set of physical objects according to some established procedure or frame of reference. In texts, depending on the culture, rows, such as in English (left to right), or in Arabic (right to left) or columns, such as in Chinese (top to bottom) are used for writing in a given direction. Rows and columns can be used together to cross reference information in texts, and a rectangular table is a convenient way of doing this. Information is frequently stored in such arrays, and conventions are required as to how information is to be read from these tables, or when and how information from such tables may be combined or otherwise manipulated. In many contexts the information stored in specific locations within a rectangular array is numerical. For example, suppose we record the number of coins of the denominations 5 cents, 10 cents, 20 cents, 50 cents, one dollar and two dollars, in that order, of spare change that several individuals Michael, Jay, Sam and Lin, also in that order, collect over a week. We might write these in an ordered list: {Michael {4, 0, 1, 3, 2, 2}; Jay {5, 1, 0, 0, 4, 2}; Sam {0, 0, 0, 4, 3, 0} and Lin {10, 4, 6, 0, 0, 1}}

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MathsWorks for Teachers Matrices

Alternatively, this information could be displayed in a table: 5 cent

10 cent

20 cent

50 cent

$ 1.00

$ 2.00

Michael

4

0

1

3

2

2

Jay

5

1

0

0

4

2

Sam

0

0

0

4

3

0

Lin

10

4

6

0

0

1

The corresponding more compact rectangular array of numbers is: R V S 4 0 1 3 2 2W S 5 1 0 0 4 2W S 0 0 0 4 3 0W S W S10 4 6 0 0 1 W T X Similarly, we might consider a business that has two outlets and sells only three products, A, B and C. We can represent the numbers of items of each product held in stock by each outlet in a table: Product A

B

C

Outlet 1

150

40

10

Outlet 2

70

20

10

The rectangular array of numbers representing current stock is 150 40 10 G. The entry in each position corresponds to the number of = 70 20 10 products of a certain type in stock at a particular outlet. This is an example of a matrix—a simple rectangular array of numbers. Particular pieces of information can be obtained by reference to the row and column in which the desired information is located. For example, the number of items of Product C at Outlet 1, which is 10, is found in the first row, third column. It is common to designate matrices by capital letters, so this matrix could be called S (the stock matrix for this context) where: S ==



12

150 40 10 G 70 20 10

chapter 2 Rectangular arrays, matrices and operations

The piece of information discussed earlier, the numbers of items of Product C at Outlet 1, can be referred to as s13, where use of the lowercase s indicates an element of the matrix S, and the subscript 13 indicates this element is found in the first row and third column of the matrix S. This matrix has two rows and three columns, and this is summarised by saying that S is a matrix of size (order or dimension) 2 by 3, or alternatively a 2 × 3 matrix. Teachers can then use students’ intuitive understanding of the practical context to provide natural definitions, and related conditions for the processes of addition, subtraction of matrices, scalar multiples of a matrix and the product of matrices. For example, if a sale is made, say, of two items of Product B from Outlet 1, this can be represented by the matrix = stock record by subtracting =

0 2 0 G, and we can easily update the 0 0 0

0 2 0 150 40 10 150 38 10 G from = G to get = G, 0 0 0 70 20 10 70 20 10

150 40 10 0 2 0 150 38 10 G-= G== G. 70 20 10 0 0 0 70 20 10 Similarly, if new stock is delivered, say for example, 10 items of Product A and 5 items of Product B are delivered to each of the outlets, this additional that is =

stock can be represented by the matrix = type can be determined by adding =

10 5 0 G, and the total stock of each 10 5 0

10 5 0 150 38 10 G to = G to give 10 5 0 70 20 10

160 43 10 150 38 10 10 5 0 160 43 10 G, that is = G+= G== G. 80 25 10 70 20 10 10 5 0 80 25 10 In this sort of context it is usual to have a periodic valuation of stock. For example, it might be the case that at the end of each month the business accountant values the stock held at each outlet. Suppose Product A costs $50 per item, Product B $30 per item and Product C $80 per item. This can be R V S50 W represented by the cost matrix S 30 W and we can easily calculate the value of SS80 WW T X the stock held by multiplication of the two matrices as follows: =



R V 50 160 43 10 S W 160 # 50 + 43 # 30 + 10 # 80 10090 G # S 30 W = = G = G== 80 # 50 + 25 # 30 + 10 # 80 80 25 10 SS WW 5550 80 T X

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It should be noted that this form of product is essentially based on a linear combination of product numbers multiplied by their corresponding cost. Students will likely wonder at some stage why multiplication of matrices is not defined in terms of the corresponding arithmetic operation between elements in the same position in two matrices of the same order, as is the case for addition and subtraction of matrices. Practical examples such as the stock context provide a basis for understanding why the linear combination definition is used. If at some time there is a 10% tax added to the cost of items, then the matrix of costs can easily be updated by multiplying the cost matrix by 100% + 10% = 110% or a factor of 1.1: R V R V R V S50 W S1.1 # 50 W S 55 W 1.1 S 30 W = S1.1 # 30 W = S 33 W SS80 WW SS1.1 # 80 WW SS88 WW T X T X T X The role of the scalar multiple can also be introduced through consideration of repeated addition, such as S + S = 2S, S + 2S = 3S, S + 3S = 4S … and so on; however, this has the limitation of restricting the scalar multiple to a whole number. It is important to distinguish between this type of multiplication, a scalar multiple of a matrix, and the product of two matrices. This simple example (see also VCAA 2005, 57–60, 149–50) demonstrates the use of matrices, and the operations of matrix addition, matrix multiplication and scalar multiplication of a matrix in a practical context where the natural, or intuitive interpretation of the processes being used motivates and models the development of matrix operations. Another context which is accessible to senior secondary school students is that of scoring for events in house or other sporting competitions. However this approach also means that the matrices used are, by definition, conformable for the operation that is to be applied, and their order is known prior to any related calculations. To deal with matrices in their own right (that is, as reified objects), independently of a particular modelling context, and to explore their general properties it is necessary to be able to describe and work with them abstractly.

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chapter 2 Rectangular arrays, matrices and operations S t u d e n t ac t i v i t y 2 . 1 a b

Use a suitable matrix product to calculate the total amount of change held by each of Michael, Jay, Sam and Lin in the given week. If the Australian–US dollar exchange rate is A$1 = US$0.76, use a suitable scalar multiple of the matrix in part a to find the equivalent value of their change in US dollars.

D e f i n i t i o n o f a m a t ri x An m × n matrix, A, is a rectangular array of numbers with m rows and n columns. We say A is of order, dimension or size, m by n and write m × n as shorthand for this. This does not mean that we wish to calculate the corresponding arithmetic product, although this will tell us the total number of elements in matrix A. Unfortunately it is not very helpful to know this as many matrices can have the same total number of elements. As in our practical example, the position of each entry, or element, in the matrix is uniquely determined by its column and row numbers. Thus, we write R V S a11 a12 f a1n W S a21 a22 f a2n W A=S W S h h h W Sam1 am2 f amn W T X where the entry in the ith row and jth column, called the (i, j) entry of A, is denoted aij. In this case the letters i and j are index variables denoting position, where i runs through 1 to m, that is 1 ≤ i ≤ m, and j runs through 1 to n, that is 1 ≤ j ≤ n. There are various notations that can be used for matrices. In this text we will use square brackets to enclose the entries of a matrix. Curve brackets are also used, however it is conventional to use only one notation in a given context. Matrices are designated using upper case letters, the entries in a matrix are identified with the corresponding lower case letter and subscript indices indicating their position; thus, we sometimes write A = [ai j ] where, as above, i is the row index and j the column index. As before, aij is the entry in the ith row and jth column of A and the ranges of i and j are understood to be those given by the order of the matrix A.

15

MathsWorks for Teachers Matrices

Two special cases of note are that a m × 1 matrix is usually called a column matrix or column vector, while a 1 × n matrix is usually called a row matrix or row vector. If a rectangular array is not available for visual display, then a matrix can be written as a list of lists of equal size, where a list is an ordered set. For example, the matrix R V S 4 0 2 W S 1 1 1 W S W S- 5 10 3 W S 0 0 3.4 W T X is the 4 × 3 matrix uniquely defined by the list {{4, 0, 2}, {1, 1, 1}, {–5, 10, 3}, {0, 0, 3.4}}. When technology is used, the data to specify a matrix is either entered into a template of a specified size (where the dimensions of the matrix needs to be specified first to obtain the desired template), or as a list of lists. Example 2.1

150 40 10 G, then A is a 2 × 3 matrix, where a21 = 70 and 70 20 10 a12 = 40 . R V S50 W If B = S 30 W, then B is a 3 × 1 column matrix, or column vector, where SS80 WW T X b11 = 50 , b21 = 30 and b 31 = 80 . If A = =

If C = 71 - 2 4 A, then C is a 1 × 3 row matrix, or row vector, where c11 = 1 , c12 =- 2 andc13 = 4 .

O p e ra t i o n s o n m a t ric e s As we have seen in the earlier practical example, matrices may be added, subtracted, multiplied by a number (scalar), or multiplied by matrices. Some of these operations are not always possible; the sizes, or orders, of the matrices involved is important, that is to say there are conditions to which two matrices need to conform for their sum, difference or product to be defined, or for them to be conformable for that operation. In practice, general computation with matrices of high order is carried out by technology, and the algorithms used by various programs to carry out these computations need the orders of the

16

chapter 2 Rectangular arrays, matrices and operations

matrices involved and definitions for the relevant processes in terms of elements and their indices. Since the various operations on matrices are defined in terms of their elements, it is important to note that these elements, and any scalars which may also be involved, are usually regarded as being drawn from some field, often the real number field, R. Thus, the operations of addition, subtraction and multiplication defined on elements of matrices are the natural operations of the relevant field. An interesting exercise for teachers to work through with students is to devise programs using basic programming constructs in a suitable high-level programming language that carry out the operations of matrix arithmetic.

A d d i t i o n a n d s u b t rac t i o n o f t w o m a t ric e s If matrices A and B are of the same size m × n then A + B is the m × n matrix with (i, j) entry aij + bij, for i = 1 to m, j = 1 to n. That is, A + B = [aij] + [bij] = [aij + bij]. In other words, we simply add all the entries in their corresponding positions throughout the matrix. Subtraction can be defined in the same way, and A – B = [aij] – [bij ] = [aij – bij]. In other words, we simply subtract all the entries in matrix B from their corresponding entries in matrix A. Example 2.2





R V R V R ]- 2g + 12 VW RS 8 10 VW S 1 - 2 W S 7 12 W S 1 + 7 S 3 5 W+ S- 3 1 W = S 3 + ]- 3g 5 + 1 W= S 0 6 W SS- 4 1 WW SS- 1 2 WW SS]- 4g + ]- 1g 1 + 2 WW SS- 5 3 WW T X T X T X T X V R V R V R R ]- 2g - 12 W S- 6 - 14 VW S 1 - 2 W S 7 12 W S 1 - 7 S 3 5 W- S- 3 1 W = S 3 - ]- 3g 5 - 1 W= S 6 4 W SS- 4 1 WW SS- 1 2 WW SS]- 4g - ]- 1g 1 - 2 WW SS- 3 - 1 WW T X T X T X T X

M u l t i p l ica t i o n b y a n u m b e r ( s ca l ar m u lt i p l e ) Given a matrix A of size m × n and a number (scalar) k, then kA is the m × n matrix with (i, j) entry kaij for i = 1 to m and j = 1 to n. That is, if A = [aij ] and k is a scalar then kA = k[aij] = [k × aij].

17

MathsWorks for Teachers Matrices Example 2.3

R V S 2 - 2W If k = 3 and A = S 3 7 W, then SS- 1 5 WW T X R V R V S 2 - 2W S 6 - 6W 3A = A + ]A + Ag = 3 S 3 7 W = S 9 21 W SS- 1 5 WW SS- 3 15 WW T X T X Note that subtraction can also be expressed in terms of a scalar multiple, with k =- 1 , and the addition operation. If A and B have the same size, then A - B = A + ]- Bg with entries the sums aij + (–bij) of the corresponding entries in A and (–B). Example 2.4



V R R V R V R V S 1 - 2 W S 7 12 W S1 + ]- 7g - 2 + ]- 12gW S- 6 - 14 W S 3 5 W- S- 3 1 W = S 3 + 3 5 + ]- 1g W = S 6 4 W SS- 4 1 WW SS- 1 2 WW SS - 4 + 1 1 + ]- 2g WW SS- 3 - 1 WW T X T X T X T X

There is a special matrix, called the zero matrix O = [oij] where oij = 0 for all i and j. For any matrix A, A - A = O . Example 2.5

If A = > i

18

1 -2 3 2 4 -1 H and B = > H, then 4 2 -1 -1 3 2 1 -2 3 2 4 -1 H+> H 4 2 -1 -1 3 2 3 2 2 1 + 2 ]- 2g + 4 3 + ]- 1g == G G== 4 + ]- 1g 2 + 3 ]- 1g + 2 3 5 1

A+B =>

chapter 2 Rectangular arrays, matrices and operations

ii

A-B => ==

1 -2 3 2 4 -1 H-> H 4 2 -1 -1 3 2

-1 -6 4 1 - 2 ]- 2g - 4 3 - ]- 1g => H G 4 - ]- 1g 2 - 3 ]- 1g - 2 5 -1 -3

iii 2A - 3B = 2 > => == iv A - A = >

1 -2 3 2 4 -1 H - 3> H 4 2 -1 -1 3 2

2 -4 6 6 12 - 3 H-> H 8 4 -2 -3 9 6

- 4 - 16 9 2 - 6 ]- 4g - 12 6 - ]- 3g H G=> ] g ] g 8 - -3 4-9 -2 - 6 11 - 5 - 8

1 -2 3 1 -2 3 0 0 0 H-> H== G 4 2 -1 4 2 -1 0 0 0

S t r u c t u r e p r o p e r t i e s o f m a t ri x a d d i t i o n a n d s ca l ar m u l t i p l ica t i o n Let A, B and C be any matrices of a given size m × n, where m and n are nonzero, then 1 the sum of any two such matrices is always defined (Closure property for addition) (Associative property for addition) 2 (A + B) + C = A + (B + C) 3 A + O = A = O + A (Identity property for addition) 4 A + (-A) = O = (-A) + A (Inverse property for addition) 5 A + B = B + A (Commutative property for addition) This collection of properties can be established by working from the general definition of matrix addition as applied to the matrices A = [aij ], B = [bij ], C = [cij ] and O = [oij ] of the same order. It may be helpful to have students undertake some general case calculations for matrices of a given order, for example 2 × 3 matrices. The results may appear to be obvious, or even trivial to students; however, care should be taken to draw to their attention that they do apply to matrices drawn from the same set of a given order (for any order) by virtue of the component-wise definition of addition of matrices, and the corresponding number properties of their elements and the corresponding number operations. This may be summarised by saying that

19

MathsWorks for Teachers Matrices

such a set of matrices forms a commutative (or abelian) group under addition. If we also consider multiplication of a matrix from this set by a scalar (scalar multiple), then for any scalars r and s the following properties also hold: (Associative property of scalar multiples) 1 r(sA) = (rs)A (Right distributive property of scalar multiple over 2 (r + s)A = rA + sA scalar addition) (Left distributive property of scalar multiple over 3 r(A + B) = rA + rB matrix addition) Again, it may be useful for students to consider the general case for matrices of a given order. Taken together, these properties show that such a set of matrices with these operations of addition and scalar multiple form what is called a vector space.

Ma t ri x m u l t i p l ica t i o n As observed from the introductory example, the product matrix A × B, or product of two matrices, also written as AB, can only be defined when the number of columns in matrix A is equal to the number of rows in matrix B. Alternatively this may be expressed by saying that the matrices A and B are conformable for the product A × B when the number of elements in the rows of matrix A is the same as the number of elements in the columns of matrix B. If A = [aij] is an m × p matrix, and B = [bij] is a p × n matrix, then AB = C = [cij] is an m × n matrix with (i, j) entry the number cij =

p

/ aik bkj

k=1 That is, the (i, j) entry of the product is obtained by multiplying each of the entries in the ith row of A by the corresponding entries in the jth column of B, and then adding all these products.

20

chapter 2 Rectangular arrays, matrices and operations Example 2.6

R V S 2 W 1 -2 3 a If A = = G and B = S- 1W, then 4 2 -1 SS 3 WW T X R V 2 1 - 2 3 S W 1 # 2 + ]- 2g # ]- 1g + 3 # 3 13 AB = = G S- 1W = = G== G 4 2 - 1 S W 4 # 2 + 2 # ]- 1g + ]- 1g # 3 3 S 3 W T X Note that we cannot form the product BA, since the number of columns of B is not equal to the number of rows of A. R V S- 1 2 W 1 -2 3 b If A = = G and B = S- 2 3 W, then 4 2 -1 SS 1 4 WW R VT X - 1 2W 1 -2 3 S AB = > H S- 2 3 W 4 2 -1 S W S 1 4W T X 1 # ]- 1g + ]- 2g # ]- 2g + 3 # 1 1 # 2 + ]- 2g # 3 + 3 # 4 == G 4 # ]- 1g + 2 # ]- 2g + ]- 1g # 1 4 # 2 + 2 # 3 + ]- 1g # 4 6 8 => H - 9 10



and

R V S- 1 2 W 1 - 2 3 BA = S- 2 3 W> H S W 4 2 -1 S 1 4W TR X V S]- 1g # 1 + 2 # 4 ]- 1g # ]- 2g + 2 # 2 ]- 1g # 3 + 2 # ]- 1gW = S]- 2g # 1 + 3 # 4 ]- 2g # ]- 2g + 3 # 2 ]- 2g # 3 + 3 # ]- 1gW SS 1 # 1 + 4 # 4 1 # ]- 2g + 4 # 2 1 # 3 + 4 # ]- 1g WW X RT V S 7 6 - 5W = S10 10 - 9 W S W S17 6 - 1 W T X

So we have the situation in Example 2.6 that AB is a 2 × 2 matrix and BA is a 3 × 3 matrix. While both products are defined, in this case they are not equal since the product matrices are of different size (order). Thus, if A and B are arbitrary matrices and the product AB is defined, it may be the case that the product BA is either not defined or, if it is defined, it may not be of the same

21

MathsWorks for Teachers Matrices

size as AB. It is important to point out this aspect of matrix multiplication to students at an early stage, and also that for matrices of a given order m × n where m and n are different, neither product will be defined. These motivate consideration of conditions under which matrix multiplication might be generally defined for a given set of matrices, and also those circumstances under which addition might be generally defined for the same set of matrices. A matrix of size m × n where m = n is called a square matrix or a n × n matrix. If A and B are two square matrices of the same size, then we can form the products AB and BA by definition, since the number of rows and columns in both matrices are equal. This also ensures that addition is defined on these matrices, and we know addition is commutative anyway. However, it is not so clear for multiplication of square matrices whether AB is the same as BA or not. Considering a particular example for two 2 × 2 matrices such as 2 3 -1 2 A => H and B = > H -1 5 2 3

2 3 -1 2 2 # ]- 1g + 3 # 2 2#2+3#3 AB = > H> H== G ] g ] g ] 1 # 1 + 5 # 2 1g # 2 + 5 # 3 -1 5 2 3 4 13 == G 11 13 ]- 1g # 2 + 2 # ]- 1g ]- 1g # 3 + 2 # 5 -1 2 2 3 and BA = > H> H== G 2 # 2 + 3 # ]- 1g 2#3+3#5 2 3 -1 5 -4 7 => H 1 21 Clearly, AB ≠ BA in this case. Inspection of a range of other examples will generally show that it is not the case that AB is the same as BA. This is also the case for square matrices of other orders, a situation which students can readily investigate using suitable technology. They will readily develop a collection of cases which show that multiplication of square matrices is, in general, not commutative. A related investigation is to see if students can identify examples, and then sets, of square matrices for which the product is commutative, for example: 3 0 1 0 G and = G 0 2 0 6 There are some important situations in which matrix products are commutative, for example when they are used to represent and compose certain types of transformations of the cartesian plane, such as rotations about the origin.

=



22

chapter 2 Rectangular arrays, matrices and operations

Teachers may also wish to consider such arguments from the general definition of matrix multiplication for the case of, for example, 2 × 2 square matrices, and consideration of the equality of two matrices. Thus: e f a b G and B = = G g h c d ae + bg af + bh then AB = > H ce + dg cf + dh if A = =



and BA = >

ea + fc eb + fd H ga + hc gb + hd

If the elements of A and B are real numbers, then AB = BA when bg = fc , af + bh = eb + fd , and ce + dg = ga + hc . Properties of matrix multiplication If A, B, and C are matrices of appropriate sizes, and k is a scalar then: (Distributive property of left multiplication 1 A(B + C) = AB + AC over addition) 2 (B + C)A = BA + CA (Distributive property of right multiplication over addition) (Associative property of multiplication) 3 (AB)C = A(BC) 4 k(AB) = (kA)B 5 AB ≠ BA in general

Z e r o a n d i d e n t i t y m a t ric e s A matrix with all entries zero is called the zero matrix (of appropriate size), and denoted O. A square matrix of size n × n with all entries zero except the diagonal entries, that is those in position (j, j) for j = 1 to n, which are all 1, is called the identity matrix of size n × n, denoted I. When the size (order) of matrices being considered is fixed, the symbols O and I can be used without ambiguity, otherwise the notation Om,n and In,n or just On (when m = n) and In can be used, as applicable.

23

MathsWorks for Teachers Matrices Example 2.7

R V S0 0 W O 3, 2 = S0 0 W is the zero matrix of size 3 × 2. SS0 0 WW T X 1 0 I2, 2 = = G is the identity matrix of size 2 × 2. 0 1 R V S1 0 0 W I 3 = S0 1 0 W is the identity matrix of size 3 × 3. SS0 0 1 WW T X Zero and identity matrices have properties similar to the numbers 0 and 1 with respect to addition and multiplication. Students should be able to convince themselves that if A is a matrix and O is the zero matrix of the same size, then A + O = A = O + A and A + (–A) = O = (–A) + A, and hence A – A = O. Similarly if A is an m × n matrix and I is the identity matrix of size n × n then AI = A, and if B is an n × m matrix then IB = B. If A is also an n × n matrix, then students should also be able to observe that AI = A = IA. Initially this may be tested by a judicious range of examples, and subsequently argued in terms of the general definitions of the relevant operations. Exploration of these operations and their conformable computations for various combinations of matrices will enable students to see matrices as rectangular arrays that can be thought of as a list of lists of the same size, as a rectangular array and as abstract reified objects in their own right. They should also be aware that while particular computations involving matrices of relatively low order can be carried out fairly readily (if somewhat tediously) by hand, more complex computations and/or computations involving matrices of higher order are generally best carried out by technology designed for this purpose. However, students should also be aware that general analysis of matrix operations is likely to involve component-wise operations with numbers using indexed sets of sums and/or products of these numbers. For example, students might be asked to show that for square matrices of a given size AO = O = OA but AB = O does not necessarily imply that A = O or B = O.

24

chapter 2 Rectangular arrays, matrices and operations S t u d e n t ac t i v i t y 2 . 2 a

1 3 2 -1 2 4 6 Given A = = G, C = = G calculate G, B = = -1 2 2 4 8 10 12 i ii iii iv v vi

2A A–B A + B AB BA AC p 4 1 G and P = = G where p and q are non-zero real numbers. Find all real q 8 6

b

Let M = =



values of the scalar k such that MP = kP.

c

11 Let A = = G . Evaluate An for n = 2 to 5 and find a general form for n > 1. 11

d e

Show that for square matrices of a given size AO = O = OA, but AB = O does not necessarily imply that A = O or B = O. Let J = =

0 -1 G and I be the identity matrix for multiplication for 2 × 2 matrices. 1 0

Show that J 2 = –I and that J 4 = I. f Explain why, if X and Y are 2 × 2 matrices, then, in general, X  2 – Y  2 ≠ (X + Y)(X – Y) and illustrate this with a suitable (counter) example. Find two matrices X and Y for which this relationship is true.

T h e t ra n s p o s e o f a m a t ri x It is a natural question to ask what happens if a matrix is written ‘the other way round’. The transpose of a matrix is obtained by interchanging its rows and columns. That is, the entries of the ith row become the entries of the ith column. So, if A = [aij] is an m × n matrix, then its transpose, denoted AT = 7aij AT = 8aTij B, is an n × m matrix, with aTij = a ji . Example 2.8

R V S1 3 W 1 2 5 If A is the 3 × 2 matrix S2 4 W, then AT is a 2 × 3 matrix = G. 3 4 8 SS5 8 WW T X

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MathsWorks for Teachers Matrices

Property of matrix transposes If A and B are matrices such that we can form the product AB, then

(AB)T = BTAT We illustrate this with an example.

Example 2.9

Let A = >

1 -2 2 -1 4 H and B = > H. 3 0 0 1 3

R V T T S 2 6 W 2 3 2 1 2 2 1 4 Then ]ABgT = f> H> Hp = > H = S- 3 - 3 W 6 - 3 12 3 0 0 1 3 S W S- 2 12 W T X R V R V S 2 0W 1 3 S 2 6 W and BT AT = S- 1 1 W> H = S- 3 - 3 W SS WW - 2 0 S W S- 2 12 W 4 3 T X T X Note that we cannot form the product AT BT, since the number of columns in AT, namely 2, is not equal to the number of rows in BT, namely 3. Most equations involving matrices can be written in two forms: the original and its equivalent using transposes. For example, the matrix equation R V 50 160 43 10 S W 10090 S G # 30 W = = G = 80 25 10 S W 5550 S80 W T X that we have seen earlier can also be written in transposed form: R V S160 80 W 650 30 80@ # S 43 25 W = 610090 5550@ SS W 10 10 W T X Commonly, matrix equations involving transformations (see Chapter 4) and transitions (see Chapter 5) may also be written in a form that is the transpose of the form given in this book. A symmetric matrix is the same as its transpose. If A = [aij] is a symmetric matrix, then aij = aji and AT = A. A symmetric matrix must be a square matrix.

26

chapter 2 Rectangular arrays, matrices and operations Example 2.10

R V 1 - 2 4W S 1 3 G and S- 2 2 0 W are symmetric matrices. = 3 2 S W S 4 0 3W T X A diagonal matrix is a square matrix that has non-zero entries only on the main diagonal. Example 2.11

R V S1 0 0 W 1 0 G and S0 - 2 0 W are diagonal matrices. = 0 2 SS W 0 0 3W T X

T h e i n v e r s e o f a m a t ri x A square n × n matrix A is said to be invertible if there is a matrix C, written as A–1, of the same size as A, such that AC = I and CA = I, where I is the identity matrix of size n × n. C is said to be the inverse matrix of A. 2 4 Not all square matrices have inverses. For example, = G does not have an 3 6 inverse. Properties of inverses It is important that students are familiar with some of the key properties of matrices and their inverses: 1 A square matrix has at most one inverse, where A–1 A = I = AA–1. 2 If A is invertible, then so is AT, and (AT) –1 = (A–1)T. 3 If A is invertible, then so is A–1, and (A–1) –1 = A. 4 If A and B are invertible matrices of the same size, then AB is invertible and (AB) –1 = B –1 A–1. These properties can be established fairly readily, and provide some good examples to students of proofs that are not lengthy, but are illustrative of important aspects of mathematical reasoning using structural properties such

27

MathsWorks for Teachers Matrices

as uniqueness, identity and inverse, definitions and the use of previously established results. Proofs of these properties are as follows: 1 If C and D are both inverses of A, then AC = I and CA = I, and AD = I and DA = I. So C = CI = C(AD) = (CA)D = ID = D, that is C = D and the inverse A–1 is unique. 2 AA–1 = I, so (AA–1)T = (A–1)TAT = I since IT = I. Hence (AT) –1 = (A–1)T. 3 AA–1 = I, so clearly A is the inverse matrix of A–1. 4 By definition (AB) –1(AB) = I. Consider (B –1 A–1)(AB), by the associative property for multiplication, this is the same as B –1(A–1 A)B, = B –1IB = B –1B = I by inverse and identity properties. So by 1, since inverses are unique, (AB) –1 = B –1 A–1. There are many uses of inverse matrices. The following are just a few. 1 The inverse of a matrix can be used for cancellation purposes in matrix equations. If A is invertible, then: i AB = AC implies that B = C (since we can multiply both sides on the left by A–1). ii BA = CA implies that B = C (since we can multiply both sides on the right by A–1). 2 4 If A is not invertible, then this is not the case. For example, let A = = G, 3 6 1 2 1 5 10 1 B == G and C = = G. 3 4 2 1 0 2

Then AB = =



and AC = =

2 4 1 2 1 14 20 10 G= G== G 3 6 3 4 2 21 30 15

2 4 5 10 1 14 20 10 G= G== G 3 6 1 0 2 21 30 15

Hence AB = AC and yet B ≠ C. 2 The inverse matrix can be used to solve a system of simultaneous linear equations with a unique solution. Consider the following system of simultaneous equations:

2x + 3y = 7 4x + y = 3

This system can be written in matrix form as AX = B, where A = =

2 3 G, 4 1

x 7 X = = G and B = = G. If A–1 exists, then we can multiply both sides of the y 3 equation on the left by A–1, and we thus have X = A–1B, and so we can find the solution by matrix multiplication.

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chapter 2 Rectangular arrays, matrices and operations

There are several ways to find the inverse of a matrix. The most useful one is via the Gauss–Jordan method (see, for example, Anton & Rorres, 2005, or Nicholson, 2003 for details). Since we shall generally only be concerned with the inverse of a 2 × 2 matrix, we will find it in a simple way. Example 2.12

Find the inverse of A = = Solution

Suppose = Then =

a b G if it exists. c d

e f G is its inverse. g h

1 0 a b e f G. G== G= 0 1 c d g h

Hence: ae + bg = 1 (i) af + bh = 0 (ii) ce + dg = 0 (iii) cf + dh = 1 (iv) Considering equation (ii), if we put f =- b and h = a , then the equation is satisfied. Similarly for equation (iii) we can put e = d and g =- c . Now consider the matrix product:

=

a b e f a b d -b ad - bc 0 G= G== G= G== G = ]ad - bcgI 0 ad - bc c d g h c d -c a

and then it is easy to see that =

a b -1 d -b 1 G = = G provided ad bc c a c d

ad - bc ! 0 , since if BA = kI, where k is a non-zero scalar, then

` k1 Bj A = I , and hence, by definition, ` k1 Bj = A- 1.

If ad – bc = 0, then the matrix A does not have an inverse. The number ad – bc is called the determinant of A and denoted det(A) or |A|. (For more information on determinants, see the references on linear algebra.)

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MathsWorks for Teachers Matrices

Inverse of a 2 × 2 matrix A 2 × 2 matrix A = =

a b G has an inverse if and only if ad – bc ≠ 0, and then c d

=



a b -1 d -b 1 G = = G ad - bc - c a c d

(The reader can check that AA–1 = A–1 A = I.) Note that an n × n matrix A has an inverse if and only if det(A) ≠ 0; see further references on linear algebra for definition and calculation of such an inverse. S t u d e n t ac t i v i t y 2 . 3 a

Find the inverse of each of the following matrices: i

b

=

1 2 G 3 4

ii =

5 3 G 3 2

iii =

2 1 G 3 -1

Use matrix inverses to solve the following system of simultaneous linear equations: 2x + 3y = 7 4x + y = 3

A p p l ica t i o n s o f m a t ric e s Fibonacci’s rabbits Suppose that newborn pairs of rabbits produce no offspring during the first month of their lives, but then produce one new pair every subsequent month. Start with F1 = 1 newborn pair in the first month and determine the number, Fr = number of pairs in the rth subsequent month, assuming that no rabbit dies. Since the newborn pair do not produce offspring in the second month, we have F2 = F1 = 1. In the third month, the original pair will produce one pair of offspring, so F3 = 2. In the fourth month, the pairs in the second month will each produce another pair, so the total will be these newborn pairs added to the number of pairs from the previous month, that is F4 = 1 + 2 = 3, or F4 = F2 + F3. Continuing in this manner, we see that F5 = (Number of pairs alive in the third month, which each produce one pair of offspring in the fifth month) + (Number of pairs alive in the fourth month)  = F3 + F4

30

chapter 2 Rectangular arrays, matrices and operations

and in general that Fr = (Number of pairs alive in month r – 1) + (Number of pairs alive in the (r – 2)th month, which each produce one pair of offspring in the rth month) = Fr – 1 + Fr – 2. Thus the sequence for the number of pairs of rabbits is 1, 1, 2, 3, 5, 8, 13, 21, 34 … which is called the Fibonacci sequence, and has the property that from the third term on, each term is the sum of the preceding two terms in the sequence. If Fr represents the rth term in this sequence, then Fr = Fr – 1 + Fr – 2 We can express this in matrix form:

>

Then writing fr = >

Fr 1 1 Fr - 1 G> H== H Fr - 1 1 0 Fr - 2

Fr 1 1 G, we find that fr = Afr – 1. H for r > 1 and A = = Fr - 1 1 0

In particular, f3 = Af2, f4 = Af3 = A2f2, f5 = Af4 = A3f2 and, in general, fr = Ar – 2f2. So elements of this sequence can be determined by finding powers of the matrix A, and multiplying by the column matrix f2.

S t u d e n t ac t i v i t y 2 . 4 Find the 29th and 30th numbers in the Fibonacci sequence.

Codes Governments, national security agencies, telecommunications providers, banks and other companies are often interested in the transmission of coded messages that are hard to decode by others, if intercepted, yet easily decoded at the receiving end. There are many interesting ways of coding a message, most of which use number theory or linear algebra. We will discuss one that is effective, especially when a large-size invertible matrix is used. At a personal level, bank accounts, credit cards, superannuation funds, computer networks, phone companies, frequent flyer schemes, electronic

31

MathsWorks for Teachers Matrices

motorway toll systems and other commercial enterprises (such as online bookstores) require PINs or passwords to be able to do a transaction over the phone, at a machine or over the Internet. Banks now warn people not to write down the PIN/password at all, but to remember it. In some cases, people record PINs/passwords disguised as phone numbers, but this is not very secure and people are encouraged not to do it. Nowadays, a person usually has many different PINs/passwords which should be changed regularly. It is impossible to remember them all. However, it would be possible to write down the information in a coded form using the methods described below. Let us start out with an invertible matrix M that is known only to the transmitting and receiving ends. For example: -3 4 M == G -1 2



Suppose we want to code the message L

E

A

V

E

N

O

W

We make a table of letters of the alphabet with the number corresponding to the position of the letter in the alphabet under it. We use 0 for an empty space. space

A

B

C

D

E

F

G

H

0

1

2

3

4

5

6

7

8

I

J

K

L

M

N

O

P

Q

9

10

11

12

13

14

15

16

17

R

S

T

U

V

W

X

Y

Z

18

19

20

21

22

23

24

25

26

We can use this table to replace each letter with the number that corresponds to the letter’s position in the alphabet. L

E

A

V

E

↑ ↓

↑ ↓

↑ ↓

↑ ↓

↑ ↓

12

5

1

22

5

32

N

O

W

↑ ↓

↑ ↓

↑ ↓

↑ ↓

↑ ↓

0

14

15

23

0

chapter 2 Rectangular arrays, matrices and operations

The message has now been converted into the sequence of numbers 12, 5, 1, 22, 5, 0, 14, 15, 23, 0, which we group as a sequence of column vectors: =



12 1 5 14 23 G, = G, = G, = G, = G 5 22 0 15 0

and multiply on the left by M: M =

- 16 - 15 1 85 14 18 23 - 69 12 5 G== G G, M = G = = G, M = G = = G, M = G = = G, M = G = = -2 -5 5 0 22 43 15 16 0 23

giving the sequence of numbers –16, –2, 85, 43, –15, –5, 18, 16, –69, 23. This is the coded message. Note that this could have been done in one step by the matrix multiplication:

M=

- 16 85 - 15 18 - 69 12 1 5 14 23 H G=> 5 22 0 15 0 - 2 43 - 5 16 23

To decode it, the recipient needs to compute M–1: -1 2 M- 1 = > 1 3 H -2 2



- 16 85 - 15 18 - 69 and multiply it by the vectors = G to get back the G, = G, = G, = G, = - 2 43 - 5 16 23 original numbers. - 16 - 15 85 1 18 14 12 5 M- 1 = G = = G, M- 1 = G = = G, M- 1 = G = = G, M- 1 = G = = G, -2 -5 5 0 43 22 16 15 - 69 23 M- 1 = G== G 23 0 - 16 85 - 15 18 - 69 12 1 5 14 23 or M- 1 > H== G - 2 43 - 5 16 23 5 22 0 15 0 Note: It is possible to use 2 × 2 matrices so that both the matrix and its inverse have integer entries—simply ensure that the determinant is equal to 1 or –1. It is also easy to extend this simple coding system to take account of alphanumeric data such as PINs.

33

MathsWorks for Teachers Matrices S t u d e n t ac t i v i t y 2 . 5 a b

Based on this approach, code the message SAVE THE WHALES using the matrix 5 3 = G. 3 2 The following message was received and was decoded using the coding matrix 5 3 M == G. 3 2 65, 42, 75, 50, 138, 87, 90, 54, 85, 54, 80, 49, 160, 99, 123, 76 Determine the original message.

SUMM A R Y

• Matrices are ordered rectangular arrays of numbers. An array with m rows of n elements (that is, with n columns) is said to be a matrix of size (order, dimension) ‘m by n’, written as m × n. Capital letters are typically used to designate matrices. • A matrix A of size m × n can be written as a list of lists: A = {{a11, a12, a13 … a1n}, {a21, a22, a23 … a2n}, {a 31, a 32, a 33 … a 3n} … {am1, am2, am3 … amn}} or using a template such as: R V S a11 a12 f a1n W S a a f a2n W A = [aij] =  S 21 22 W h h W S h Sam1 am2 f amn W T X where aij is the element in the ith row and the jth column, and 1 ≤ i ≤ m and 1 ≤ j ≤ n. • Matrices can be added (are conformable for addition) if they are of the same size. The sum of two matrices is obtained by adding the elements in corresponding positions, that is A + B = [aij] + [bij] = [aij + bij] for all 1 ≤ i ≤ m and 1 ≤ j ≤ n. For any matrices A, B and C of the same size, A + B = B + A and (A + B) + C = A + (B + C). • The zero matrix of size m × n is defined by Om, n = [oij ] where oij = 0 for all 1 ≤ i ≤ m and 1 ≤ j ≤ n, and is often written as O when the order is known (fixed) in a given context and there is no ambiguity.

34

chapter 2 Rectangular arrays, matrices and operations

SUMM A R Y (Cont.)

• If A is a matrix of size m × n and k is a scalar then kA = k[aij] = [kaij] for all 1 ≤ i ≤ m and 1 ≤ j ≤ n. In particular –1A = [–aij] = –A. For any matrix A and the corresponding zero matrix O, A + O = A = O + A, A + (–A) = O = (–A) + A and A – A = O. For any matrix A, and scalars r and s, r(sA) = (rs)A. • If A is an m × n matrix and B is a p × q matrix then the product C = AB is defined when n = p and is a matrix of size m × q. Matrix p

•

•

•

•

•

multiplication is defined ‘row by column’ where cij = / aik bkj for k=1 all 1 ≤ i ≤ m and 1 ≤ j ≤ q. In general, AB ≠ BA. If A, B and C are matrices and k is a scalar then (AB)C = A(BC) and k (AB) = (kA)B, provided the relevant matrix products are defined. The n × n identity matrix, written In, n = In or just I when the order is given and no ambiguity arises, is defined as the matrix with 1 for each element along its leading diagonal (top left to bottom right) and 0 for all other elements. For any square matrix A of a given order and the corresponding identity matrix I, AI = A = IA. If A, B and C are square matrices of a given size, and r and s are scalars, then A(B + C) = AB + AC, (B + C)A = BA + CA, (r + s)A = rA + sA and r(A + B) = rA + rB. The transpose of a matrix A = [aij] is the matrix AT = [aij]T = 8aTij B where aTij  = aji. If the product AB of two matrices is defined, then (AB)T = BTAT. The inverse of a square matrix A is a square matrix A–1 such that AA–1 = A–1 A = I, the identity matrix of the same size as A. If AB is the product of two square matrices of the same size, then (AB) –1 = B –1 A–1. d -b a b 1 For a 2 × 2 matrix A = = G, A- 1 = = G, provided ad - bc - c a c d ad – bc ≠ 0. (ad – bc is called the determinant of matrix A, written det(A) or |A|.)

35

MathsWorks for Teachers Matrices

References Anton, H & Rorres, C 2005, Elementary linear algebra (applications version), 9th edn, John Wiley and Sons, New York. Cirrito, F (ed.) 1999, Mathematics higher level (core), 2nd edn, IBID Press, Victoria. Hill, RO Jr 1996, Elementary linear algebra with applications, 3rd edn, Saunders College, Philadelphia. Lipschutz, S & Lipson, M 2000, Schaum’s outlines of linear algebra, 3rd edn, McGraw-Hill, New York. Nicholson, KW 2001, Elementary linear algebra, 1st edn, McGraw-Hill Ryerson, Whitby, ON. Nicholson, KW 2003, Linear algebra with applications, 4th edn, McGraw-Hill Ryerson, Whitby, ON. Poole, D 2006, Linear algebra: A modern introduction, 2nd edn, Thomson Brooks/ Cole, California. Sadler, AJ & Thorning, DWS 1996, Understanding pure mathematics, Oxford University Press, Oxford. Victorian Curriculum and Assessment Authority (VCAA) 2005, VCE Mathematics study design, VCAA, East Melbourne. Wheal, M 2003, Matrices: Mathematical models for organising and manipulating information, 2nd edn, Australian Association of Mathematics Teachers, Adelaide.

Websites http://wims.unice.fr/wims/en_tool~linear~matrix.html This website provides a matrix calculator. http://en.wikipedia.org/wiki/Matrix_(mathematics) This website gives a concise introduction to matrices and matrix arithmetic, and has links to other resources and references. http://www.sosmath.com/matrix/matrix.html This site has some notes on basic matrix concepts and operations at quite a simple level.

36

C ha p t e r

3

S o lv i n g s ys t e m s o f s i m u l t a n e o u s l i n e ar eq u at i o n s From the middle years of secondary schooling, students become familiar with linear relations of the form ‘a certain number of x s added to a certain number of y s are equal to a given number’, such as 2x + 3y = 24. Usually this is done by considering a table of whole number ordered pairs of values that satisfy the relation, and then plotting the corresponding points on a set of cartesian axes. This is then typically extrapolated (by an implicit continuity assumption) to consideration of the continuous straight line on which these points lie. Thus students learn to draw the graph, part of which is shown in Figure 3.1, of such y 15 10 5 –20

–15

–10

–5

O

5

10

15

x

20

–5 –10 –15 –20 Figure 3.1: Part of the graph of the linear relation 2x + 3y = 24

37

MathsWorks for Teachers Matrices

linear relations, by identification of their axis intercepts and drawing in the line containing these two points. The corresponding working might go something like: when x = 0, 3y = 24 and so y = 8, hence (0, 8) are the coordinates of the vertical, or y-axis, intercept. Similarly, when y = 0, 2x = 24 and so x = 12, hence (12, 0) are the coordinates of the horizontal, or x-axis, intercept. Students will also learn how to identify the gradient of the straight line from this form, the graph, or possibly by re-writing it in the function form -2 y = 3 x + 8. A single linear equation in two variables ax + by = k, where a, b and k are real constants, is used to define the linear relation corresponding to the set of points {(x, y): ax + by = k, x, y ∈ R} and this set of points can be used to draw the graph of a straight line in the cartesian plane R2. Any ordered pair (x, y) that satisfies this equation is a point on the graph of the straight line. If we are asked to consider the set of points which satisfy both this relation and another relation of the same kind together, then we have a set of two simultaneous linear equations in the variables x and y. Given that each relation corresponds to the graph of a straight line in R2, we can interpret this geometrically to see that there are three possibilities: • The simultaneous linear equations corresponding to the rules of these relations have a unique solution, and their graphs have a unique point of intersection. • The simultaneous linear equations corresponding to the rules of these relations have no solution, and their graphs are parallel and distinct straight lines. • The simultaneous linear equations corresponding to the rules of these relations have infinitely many solutions, the equations represent the same relation and their graphs are the same straight line. As a related activity students could be asked to identify the rules of several other linear relations that correspond to each of these cases with respect to 2x + 3y = 24, and verify these by using technology to draw the corresponding graphs. In such an activity, students are not considering the relationship in terms of its component parts, but as an object in its own right: each relation has a graph, and this graph may or may not have certain properties with respect to the given relation 2x + 3y = 24 and its graph. In the first and last cases described above, the system of simultaneous linear equations is said to be consistent, in the second case it is said to be inconsistent. It is likely that a set of three or more arbitrarily selected

38

chapter 3 Solving systems of simultaneous linear equations

simultaneous linear equations in two variables will be inconsistent, however this is not always the case. For example the set of simultaneous linear equations {3x + 2y = 5, x – y = 0, –x + 4y = 3} has the unique solution (1, 1). Indeed, teachers can ask students to form systems of several simultaneous linear equations in two variables that are satisfied by a given ordered pair (m, n), simply by choosing the constant k to be am + bn for each combination a and b of coefficients for x and y in ax + by = k. Students will likely be familiar with solving simultaneous systems of two linear equations in two variables, using graphical, numerical and algebraic approaches, from their work in the middle years of secondary mathematics. The linear equations involved will have been alternatively presented in the forms y = mx + c, ax + by = k and Ax + By + C = 0, and students should be able to convert algebraically between these forms. They should also be aware that the first form only applies to linear relations that are functions; that is, it is not possible to use this form to describe linear relations such as {(x, y): x = 6, y ∈ R}. It is perhaps useful in this context to explicitly point out how this relates to the coordinate specification of the position of a point in the cartesian plane. For example, as shown in Figure 3.2, the point with coordinates (4, 7) corresponds to the solution of the pair of simultaneous linear equations x = 4, or 1x + 0y = 4, and y = 7, or 0x + 1y = 7: y 15 10

y=7

(4, 7)

5 –4

–2

O

–5

2

4

6

8

x

10

x=4

–10 –15 –20 Figure 3.2: Parts of the graphs of the lines with equations x = 4 and y = 7 showing their point of intersection at (4, 7)

39

MathsWorks for Teachers Matrices

Practical and theoretical problems in many areas can be formulated in terms of solving a system of simultaneous linear equations. A linear equation in two variables is a special case of an equation of the form f(x, y) = k, where k is a real constant and f(x, y) has the form ax + by for real constants a and b. For example, f(x, y) = 3x – 2y = 10 is a (linear) equation where k = 10 and a and b are 3 and –2 respectively. Students would be familiar with the use of this form to specify the rule of a linear function in the cartesian plane, or R2, and its corresponding straight line graph, with gradient 23 and axis-intercepts j at (0, –5) and ` 10 3 , 0 . If g(x, y) = l also has the same form, for example 4x + 5y = –3, then we have a pair of simultaneous linear equations {f(x, y) = k, g(x, y) = l} or {3x – 2y = 10, 4x + 5y = –3}. The solution to this pair of simultaneous linear equations will be the set of all ordered pairs (x, y) - 49 j that satisfy both equations, in this case the ordered pair ` 44 23 , 23 . The solution can be readily found using a computer algebra system (CAS), such as Derive: SOLVE([3.x – 2.y = 10, 4.x + 5.y = –3], [x, y])



[x =



44 49 23 ^ y = – 23

]

This ordered pair represents the point of intersection of the corresponding straight line graphs in the cartesian plane, as shown in Figure 3.3. y 6 4 2

–2

–1

O

1

2

3

4

–2 –4 –6

Figure 3.3: Graph of two linear functions 3x – 2y = 10, 4x + 5y = –3 showing their point of intersection

40

x

chapter 3 Solving systems of simultaneous linear equations

We can similarly define a linear equation in three variables as a special case of an equation of the form f(x, y, z) = k, where k is a real constant and f(x, y, z) has the form ax + by + cz for real constants a, b and c. For example, if f(x, y, z) = 3x – 2y + z, then 3x – 2y + z = 0 is a (linear) equation where k = 0 and a, b and c are 3, –2 and 1 respectively. Some students will have worked with this form as representing the equation of a plane in three-dimensional cartesian space, or R3. Any point with coordinates (x, y, z) that satisfy this relation is part of the corresponding plane. Again, this single equation has infinitely many solutions—all the points in the plane that it defines. Technologies such as CAS are useful tools to assist in graphically representing three-dimensional shapes. We can similarly form sets of simultaneous linear equations involving three variables, with two or more equations. For example, the intersection of 3x – 2y + z = 0 with x – y – z = 10, corresponds to the solution set:

SOLVE([3.x – 2.y + z = 0, x – y – z = 10], [x, y])



[x = –3.z – 20 ^ y = –2(2.z + 15)]

This is a parametric solution in terms of the variable z, where for each real value of z the corresponding values of x and y are given in terms of z. Thus, the solution set is {(–3z – 20, –2(2z + 15), z): z ∈ R}. Each value of z generates the coordinates of a point in R3, represented by an ordered triple, and these points all lie on the straight line formed in the three dimensional space, R3, where the two planes intersect, as shown in part in Figure 3.4.

5 z –5 25 –25 y x 25

–25

Figure 3.4: Graph of parts of 3x – 2y + z = 0 and x – y – z = 10 showing intersection

41

MathsWorks for Teachers Matrices

This process can be extended to systems defined by sets of simultaneous linear equations in as many variables as we wish, although it is not then possible to provide simple geometric interpretations for more than three variables. CAS, and other technologies, can be used to solve systems of simultaneous linear equations directly in ‘black-box’ mode (that is, providing results without detailing intermediate steps of calculation). In this text the use of such technologies is intended to elucidate the processes that are employed in applying algorithms to obtain solutions. While mental and by hand computation is an important part of the experience of developing understanding of key concepts, skills and processes, particularly in less complicated illustrative cases, CAS and other technologies have been developed to facilitate problem solving and analysis of the behaviour of mathematical systems by application of their functionality. They are essential tools to employ in contexts where many computations are required to be carried out quickly and reliably, such as with matrices of large size (order). Although judgments about when, where and how it is appropriate to employ enabling technology (or not) will naturally vary with practical and philosophical considerations in context, individuals will ultimately make their own decisions on this matter where they have a choice. Teachers may find it useful to discuss with students a range of perspectives and considerations on this issue, including their own views and the rationales for these views. Matrices provide a natural model for representations and manipulation of systems of simultaneous linear equations. For example, the linear form of f(x, y, z) = k corresponds to multiplying the variables x, y and z by their respective coefficients a, b and c, and making the sum of these values equal to k, that is, ax + by + cz = k. This is precisely the same as matrix row by column multiplication: R V Sx W 7a b c A # S y W = 5 k ? SS z WW T X Moreover, this extends naturally to a system of simultaneous linear equations, because for matrices such multiplication is defined for each combination of rows and columns where the matrices involved are conformable for multiplication. That is, the equations involved have the same number of variables, even if some of their coefficients are zero. For example, the equations 3x – 2y + z = 0 and x – z = 4 can be represented as the two

42

chapter 3 Solving systems of simultaneous linear equations

R V R V Sx W Sx W matrix equations, 7 3 - 2 1A # Sy W = 5 0 ? and 71 0 - 1A # Sy W = 5 4 ? SS WW SS WW z z T X T X respectively, or simultaneously via the single matrix equation R V Sx W 0 3 -2 1 H # Sy W = = G. > 1 0 -1 S W 4 Sz W T X S t u d e n t ac t i v i t y 3 . 1 a b c d

Write down a system of two simultaneous linear equations in two variables that has (–5, 6) as its unique solution. Write down a system of two simultaneous linear equations in two variables that has (–5, 6) as one of its many solutions. Write down a system of two simultaneous linear equations in three variables that have (0, 0, 0) and (1, 1, 1) as solutions. Use the Solve functionality of a CAS, or a like functionality of other suitable technology, to find the intersection of 3x – 2y + z = 0 and x – y – z = 10, and express the solution set in terms of y.

S o l v i n g s y s t e m s o f s i m u l t a n e o u s l i n e ar e q u a t i o n s u s i n g m a t ri x i n v e r s e If a system of n simultaneous linear equations in n variables is consistent and has a unique solution, then square matrices and their inverse matrices may be used to find this unique solution. We will begin by considering two simple examples, give some geometric interpretations and then introduce some general notation. To start with, it is important for students to see how a familiar set of two simultaneous linear equations in two variables may be written as an array, as is often the case for by hand techniques, and subsequently as a single matrix equation.

43

MathsWorks for Teachers Matrices Example 3.1

Consider the system of two simultaneous linear equations in two variables (often called unknowns in this context), x and y, defined by requiring two numbers x and y to have a difference of 1 and a sum of 3. This is given by {x – y = 1, x + y = 3} and can be written in a rectangular array form, with corresponding variables vertically aligned x-y = 1 x 1 -1 1 as * 4 or in matrix form as = G # = G = = G. y 1 1 3 x+y = 3 In this case the solution of x = 2 and y = 1 can be readily identified by inspection. However, this will not always be the case. The system of three simultaneous linear equations in three variables, x, y and z given by {x – y – z = 0, 6x + 4y = 20, -4y + 2z = 10} can be written in a rectangular array form, with corresponding variables vertically aligned as R V R V R V Z x - y - z = 0_ S1 - 1 - 1W S x W S 0 W ] b [ 6x + 4y + 0z = 20` or in matrix form as S6 4 0 W# Sy W = S20 W. SS0 - 4 2 WW SS z WW SS10 WW ] 0x - 4y + 2z = 10b \ a T X T X T X Other more complicated systems of n equations in n variables can also be likewise represented. R V 1 - 1 - 1W S 1 -1 The matrices = G and S6 4 0 W are commonly called the coefficient 1 1 S W S0 - 4 2 W T R V X R V 0 S W Sx W x 1 S W matrices, the matrices = G and 20 the constant matrices and = G and Sy W the y 3 SS10 WW SS z WW T X T X matrices of the variables or unknowns. Students should be encouraged to note that the matrices involved hold systematic information about the system of equations. Each column of the coefficient matrix contains the coefficients of one variable, one coefficient from each of the equations. The variables are thus ordered according to which column of the matrix they correspond to. If we write A for the coefficient matrix, B for the constant matrix and X for the matrix of unknowns, then each of the above systems of simultaneous equations (and any other like system) can be represented in the same form by a single matrix equation

44

chapter 3 Solving systems of simultaneous linear equations

AX = B. This equation can be solved for X by left multiplying both sides of the matrix equation by A–1 and applying matrix algebra: AX = B A–1(AX) = A–1B (A–1 A)X = A–1B IX = A–1B X = A–1B

(multiplying by A–1 on left) (by associativity of matrix multiplication) (since (A–1 A) = I) (since IX = X)

For the system of two simultaneous linear equations in two unknowns this gives: R V S 1 1W 1 x 1 - 1 -1 1 2 G = G = S 21 12 W= G = = G = G== y 1 1 3 1 SW3 S 2 2W T X So x = 2 and y = 1 is the (simultaneous) solution to both the equations, as expected. Using the matrix inverse, both values are obtained at the same time, unlike other techniques in which the values are determined successively. Geometrically, each equation corresponds to a straight line in the cartesian plane, and they intersect at the point with coordinates (2, 1), as shown in Figure 3.5. y 3

x+y=3

2 1 –1

O

1

2

3

4

x

5

–1 –2 x–y=1

–3

Figure 3.5: Graphs of x – y = 1 and x + y = 3 and their point of intersection

45

MathsWorks for Teachers Matrices

For the system of three simultaneous linear equations in three unknowns this gives: R V R V 2 3 1 W 40 W S S R V R1 - 1 - 1V- 1 R V 11 22 11 WR 0 V S 11 W W S0W S Sx W S S W Sy W = S6 4 0 W S20 W = S- 3 1 - 3 WS20 W = S- 5 W W S W S 11 22 22 WSS WW S 11 W SS WW SS S 45 W 0 - 4 2 W S10 W S 6 1 z 5 W 10 T X T X T X S- 11 11 22 WT X S 11 W T X T X 40 5 45 So x = 11 , y =- 11 and z = 11 is the simultaneous solution to all three linear equations, and this is not really evident ‘by inspection’. Geometrically, each equation corresponds to a plane in the three dimensional space of R3, and 40 - 5 45 , 11 , 11 i as shown in these planes intersect at the point with coordinates _ 11 Figure 3.6.

5

z

–5 5 5

x y –5

–5

Figure 3.6: Graphs of x – y – z = 0, 6x – 4y = 20 and –4y + 2z = 10 and their point of intersection

46

chapter 3 Solving systems of simultaneous linear equations

These are called consistent systems, as there is a solution. In both cases the solution is also unique. This simple matrix method is ideal if there is a unique solution. However, in many cases there may be no solution or infinitely many solutions, and in these cases the coefficient matrix does not have an inverse. Intuitively students should be able to discuss and identify geometric interpretations of possible intersections of graphs of straight lines (using rulers) and planes (using sheets of paper) corresponding to the twodimensional and three-dimensional cases respectively. If two students write down the equations of two straight lines of the form ax + by = c independently and then compare, there are three possibilities: • They correspond to distinct lines with different gradients which have a unique point of intersection, for example {3x + 2y = 4, –x + y = 0}. • They correspond to distinct lines with the same gradient which have no points of intersection, that is they are parallel with different axis intercepts, for example, {3x + 2y = 4, 3x + 2y = 0}. • They correspond to the same line with the same gradient and have infinitely many points of intersection, that is they are parallel with the same axis intercepts, for example {3x + 2y = 4, 1.5x + y = 2}. If three students write down the equations of three planes of the form ax + by +cz = d independently and then compare, there are also several possibilities: • The three planes are distinct and intersect in a unique point. • The three planes are distinct and intersect in a line (like a three-page book). • The three planes are identical and intersect in infinitely many points that form a single common plane. • The three planes do not intersect all together. (There are several ways in which this might occur geometrically.) Students should be able to discuss the idea that, as a single equation in three variables of the form ax + by + cz = d represents a single plane in R3, then a system of two simultaneous linear equations in three variables could have zero or infinitely many solutions, the former because the planes are parallel but distinct, the latter either because the planes are identical or because they define a line in R3 by their set of intersection points. Student exploration of these possibilities will be aided by access to CAS with twodimensional and three-dimensional graphing functionalities.

47

MathsWorks for Teachers Matrices Example 3.2

Consider the following systems of two simultaneous linear equations in x+y = 4 - 6x + 2y = - 8 the two variables x and y: * 4. 4 and * 3x - y = 4 x+y = 2 Their coefficient matrices are =

-6 2 1 1 G and = G respectively, and 3 -1 1 1

neither of these has an inverse. That is, they are both singular matrices as their determinants are both equal to zero. For the first system, the corresponding lines are parallel and do not intersect, as shown in Figure 3.7, so there are no solutions. This system is said to be inconsistent. y 5 4 3

x+y=4

2 1 –2

–1

O

x+y=2 x 1

2

3

4

5

–1 –2 –3 Figure 3.7: Graphs of x + y = 2 and x + y = 4, parallel straight lines with no points of intersection

For the second system, the corresponding lines are in fact the same line, as shown in Figure 3.8, hence each point on the line is a solution to the system. This system is consistent, but with infinitely many solutions.

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chapter 3 Solving systems of simultaneous linear equations

y 6 4 2

–2

O

x 2

4

–2 –4 –6 Figure 3.8: Graphs of 3x – y = 4 and –6x + 2y = –8, identical straight lines with infinitely many points of intersection

If we use y = k as the free variable, a set of solutions, or solution set, can be written parametrically in the form $` 4 +3 k , kj: k ! R. .There are infinitely many other ways of writing the solution set for this system. For example, another solution set, using x = t as the free variable, is {(t, 3t – 4): t ! R}. Each value of k or t, as applicable, generates the coordinates of a solution point.

S t u d e n t ac t i v i t y 3 . 2 a

b

c d

The system of linear equations {ax + by = 0, cx + dy = 0}, where a, b, c and d are real constants, is called a homogeneous system. Show that for this system there is either a unique solution or infinitely many solutions. Describe relationships between real constants a, b, c, d, e and f for which the system of simultaneous linear equations {ax + by = e, cx + dy = f} has: – a unique solution – no solution – infinitely many solutions A system of equations has solution set {(3t + 1, 2t – 1): t ∈ R}. Find the corresponding cartesian equation. - 6x + 2y = - 8 Write the solution set to the system of simultaneous equations ) 3 3x - y = 4 in two ways that are different from the ways given above.

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MathsWorks for Teachers Matrices

Although we cannot use the inverse of the coefficient matrix to solve the above systems, we can use another technique called Gaussian elimination to solve such systems. This is a generalisation, of the process of eliminating variables, carried out systematically, and represented in matrix form. This algorithm can be applied in cases where there are not the same number of equations and variables—an important technical generalisation.

T h e m e t h o d o f Ga u s s ia n e l i m i n a t i o n We have seen above that a system of simultaneous linear equations can be written in matrix form. While the method involving the inverse of the matrix of coefficients can be used when the system has the same number of equations as variables within an equation, and the required inverse exists (that is, the system is consistent and has an unique solution), it would be useful to have a more general method of analysis. Such a method should: • enable us to determine whether a system is consistent or not, irrespective of its order • determine the solutions to the system, involving parametric forms as applicable • generalise existing methods for known simple cases. Such a method exists, and is called Gaussian elimination. It is a numerical method, involving only the coefficients and constants of the system of equations, and can be effectively represented using matrices. This method underpins the processes used in technology such as CAS for solving systems of simultaneous linear equations and related operations. To do this we simply use the coefficient matrix and the matrix of constants as discussed previously, and place them side by side to form a new matrix called the augmented matrix for the system of simultaneous linear equations. Example 3.3

x-y = 1 1 -1 4 has augmented matrix = 1 1 x+y = 3 Z x - y - z = 0_ ] b The system [ 6x + 4y + 0z = 20` has augmented matrix ] 0x - 4y + 2z = 10b \ a The system *

50

1 G. 3 R V S1 - 1 - 1 0 W S6 4 0 20 W. SS0 - 4 2 10 WW T X

chapter 3 Solving systems of simultaneous linear equations

In general, the matrix equation AX = B, where A is an m × n coefficient matrix, X is the n × 1 matrix of variables and B is the n × 1 matrix of constants, gives rise to the m × (n + 1) (that is, one additional column) augmented matrix: R V S a11 a12 f a1n b1 W S a21 a22 f a2n b2 W W 6 A ; B@ = S S h hW SSam1 am2 f amn bn WW T X The important discussion is to lead from this form, which is only a representation of all the coefficients and constants of the original system, to another form which enables us to read off the solutions for x1 through to xn. The processes by which we move from one form to another must ensure that these forms are equivalent, where two systems of simultaneous linear equations are said to be equivalent if each has the same set of solutions. The idea of Gaussian elimination is to solve a system of simultaneous linear equations by writing a sequence of systems, each one equivalent to the previous system. Then each of these systems has the same set of solutions as the original one. The aim is to end up with a system that is easy to solve, such as one in what is called triangular form. Instead of writing the system of equations out each time, we simply write the corresponding augmented matrix, since the natural ordering of the matrix takes care of ‘tracking’ what happens to the coefficients of the variables. To do this, only a certain type of operation, called an elementary operation, can routinely be performed on systems of simultaneous linear equations to produce equivalent systems. These are the operations we would use in solving such a system by hand. 1 Interchange: Any two equations can be interchanged. 2 Scaling: We can multiply an equation by a non-zero constant. 3 Elimination: We can add a constant multiple of one equation to another equation. In practice, operations 2 and 3 are often applied in conjunction to add or subtract a multiple of one equation from a multiple of another equation, usually to ‘eliminate’ a variable from one equation. Elementary operations performed on a system of simultaneous equations produce corresponding manipulations of the rows of the augmented matrix. Hence, either by hand or using technology, we manipulate the rows of the

51

MathsWorks for Teachers Matrices

augmented matrix rather than the equations. These row operations have the same effect as the operation on equations, where the equations have been written as a rectangular array with the coefficients of the variables and the constants vertically aligned. The following are the corresponding elementary row operations for a matrix: 1 Interchange: Interchange any two rows in their entirety. 2 Scaling: Multiply any row by a non-zero constant. 3 Elimination: Add a constant multiple of one row to another row. A matrix is said to be in (row) echelon form if it satisfies the following two conditions: 1 If there are any zero rows, they are at the bottom of the matrix. 2 The first non-zero entry in each non-zero row (called the leading entry or pivot) is to the right of the pivots in the rows above it. Matrices which are in row echelon form have a ‘staircase’ appearance: R V S0 * * * * W S0 0 * * * W S0 0 0 0 * W S W S0 0 0 0 0 W T X Example 3.4

The following matrices are in echelon form: R V R V 2 5 W S2 3 5 W 1 2 S 3 4 5 0 0 2 3 G, S0 7 W, S0 0 4 W, = G, = G = 0 3 SS 0 2 1 0 0 0 4 0 0 WW SS0 0 0 WW T X T X This idea can be extended further: a matrix is said to be in reduced (row) echelon form if it is in echelon form and also: 3 Each pivot is 1. 4 Each pivot is the only non-zero entry in its column. Example 3.5

The following matrices are in reduced row echelon form: R V R V 1 0 W S1 3 0 W 1 0 S 1 0 5 0 0 1 0 G, S0 1 W, S0 0 1 W, = G, = G = 0 1 SS 0 1 1 0 0 0 1 W S W W S W 0 0 0 0 0 T X T X

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chapter 3 Solving systems of simultaneous linear equations

Two matrices A and B are said to be equivalent if one can be obtained from the other by a finite sequence of elementary row operations, and we write A ~ B to denote this equivalence. Gaussian elimination is a procedure for bringing a matrix to its equivalent row echelon form, and is described in steps 1–4 of the Gauss–Jordan algorithm given below. At this stage the corresponding system of simultaneous linear equations is in triangular form and could be solved by back-substitution. The Gauss–Jordan algorithm, described in Table 3.1 is an extension of Gaussian elimination which brings the matrix to its equivalent reduced row echelon form from which the solution (if there is one) can be directly written down. Table 3.1: Gauss–Jordan algorithm

Gauss–Jordan algorithm

Step 1

Identify the leftmost non-zero column.

Step 2

If the first row has a zero in the column of Step 1, interchange it with one that has a non-zero entry in the same column.

Step 3

Obtain zeros below the leading entry (also called a pivot) by adding suitable multiples of the top row to the rows below it.

Step 4

Cover the top row and repeat the same process with the leftover sub-matrix and starting at step 1. Repeat this process with each row. (At this stage the matrix is in echelon form.)

Step 5

Start with the last non-zero row, work upwards. For each row, obtain a leading 1 (by dividing by the value of the pivot) and introduce zeros above it by adding suitable multiples of the row with the leading 1 to the corresponding rows.

Example 3.6

Use the Gauss–Jordan algorithm to solve the system of simultaneous linear equations formed by requiring two numbers x and y to have a difference of 1 and a sum of 3:

*

x-y = 1 4 x+y = 3

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MathsWorks for Teachers Matrices

As noted earlier, in this case the solution of x = 2 and y = 1 can readily be obtained by inspection, however this will not always be the case. A simple example such as this enables students to attend to the process being illustrated rather than focus on the manipulations involved. Solution

This system has the corresponding augmented matrix =

1 -1 1 G. 1 1 3

• The leftmost non-zero column is the first column (Step 1). • The top entry in this column is non-zero, so proceed to Step 3 of the algorithm (Step 2).) • The new second row will be the old second row minus the first row (Step 3). 1 -1 1 • The resulting matrix is = G, which is in row echelon form. 0 2 2

(This corresponds to the system of simultaneous linear equations x-y = 1 * 4 , which is in a triangular form, and could easily be solved 2y = 2 by back-substitution. The last equation is 2y = 2, and so y = 1. Substitute this value for y in the first equation, and solve for x, which gives x – 1 = 1, and so x = 2, hence the solution of the system is (2, 1).) • Now we must use Step 5 to convert the matrix to reduced row echelon form. First, we need to turn the leading entry in the second row, into a leading ‘1’. So divide the second row by 2 to obtain =



1 -1 1 G. To convert this to reduced row echelon form, we need to 0 1 1

turn the entry in the first row, second column to 0. This can be done by writing a new first row which is equal to the second row added to the first row to obtain =



1 0 2 G. 0 1 1

The matrix is now in reduced row echelon form, corresponding to the x=2 equivalent system of simultaneous linear equations * 4, which is y=1 the solution, as noted earlier.

54

chapter 3 Solving systems of simultaneous linear equations Example 3.7

Use the Gauss–Jordan algorithm to solve the following system of simultaneous linear equations: Z x - y - z = 0_ ] b [ 6x + 4y + 0z = 20` ] 0x - 4y + 2z = 10b \ a Solution

This system has the corresponding augmented matrix: R V S1 - 1 - 1 0 W S6 4 0 20 W SS0 - 4 2 10 WW T X The leftmost non-zero column is again the first column (Step 1) and the top entry in this column is non-zero (Step 2), so we proceed to Step 3 of the algorithm. The new second row will be the old second row with six times the first row subtracted from it, and since the element of the first column in the third row is already zero, we do not need to do anything to the third row at this stage. The equivalent matrix is: R V S1 - 1 - 1 0 W S0 10 6 20 W SS0 - 4 2 10 WW T X Now we are at Step 4 of the elimination process. If we cover the top row, the leftmost non-zero column is now the second column (Step 1) and the top entry is non-zero (Step 2), so the new third row will be the 4 previous third row with 10  × the second row added to it: R V S1 - 1 - 1 0 W S0 10 6 20 W SS0 0 4.4 18 WW T X This matrix is now in echelon form, and we could use backsubstitution to solve the corresponding system of simultaneous linear Zx - y - z = 0_ ] b equations [ 10y + 6z = 20`. ] 4.4z = 18b \ a

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MathsWorks for Teachers Matrices

Now we are at Step 5. We begin by dividing the last row by 4.4 = 22 5, to obtain a leading 1. The matrix is then: R V S1 - 1 - 1 0 W S0 10 6 20 W S W S0 0 1 45 W S 11 W T X We must next turn the first two numbers in the third column to 0 by using elementary row operations. The new first row will be the previous first row + the third row, and the new second row will be the previous second row with 6 × the third row subtracted from it. The new matrix will be: R V S1 - 1 0 45 W 11 W S 50 W S S0 10 0 - 11 W S 45 W S0 0 1 11 W T X Next, we need to divide the second row by 10 to obtain a leading 1: R V S1 - 1 0 45 W 11 W S 5W S S0 1 0 - 11 W S 45 W S0 0 1 11 W T X Finally, to get the matrix into reduced row echelon form, we need to obtain a 0 in the first row, second column position. This we can do by adding the second row to the first row and replacing the previous first row with this. R V S1 0 0 40 W 11 W S 5W S S0 1 0 - 11 W S 45 W S0 0 1 11 W T X The final equivalent system of simultaneous linear equations is: Z 40 _ ] x = 11 b ] b [ y = -115` ] b ] z = 45 b 11 \ a This is the required solution.

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chapter 3 Solving systems of simultaneous linear equations

Students may well inquire what happens when this algorithm is applied to a system of simultaneous linear equations that corresponds to a pair of parallel lines or identical lines in the cartesian plane. Example 3.8

Use the Gauss–Jordan algorithm to solve this system of simultaneous linear equations: *



x+y = 4 4 x+y = 2

Solution

This system corresponds to a pair of distinct parallel lines (same gradient, different intercepts). The corresponding augmented matrix for this system is =

1 1 4 G. 1 1 2

The first step in the Gauss–Jordan algorithm is to replace the second row with the previous second row minus the first row, to obtain

=

1 1 4 G. 0 0 -2

(The system is now in triangular form, and we can see that the last row corresponds to the equation 0x + 0y = –2, which is impossible. Hence there are no solutions to this system.) Next, we divide the elements in the second row by –2 to obtain 1 1 4 G, and finally we subtract four times the second row from the 0 0 1 1 1 0 first row to obtain = G. Now we see that the last equation is 0 0 1 =

0x + 0y = 1, which is impossible, and so there are no solutions to this system of equations. The above is typical of the result of elimination when there are no solutions. The last non-zero row of the augmented matrix will have zeroes everywhere except in the right-most position.

57

MathsWorks for Teachers Matrices Example 3.9

Use the Gauss–Jordan algorithm to solve the system of simultaneous linear equations: - 6x + 2y = - 8 ) 3 3x - y = 4

Solution

This system corresponds to a pair of identical parallel lines (same gradient, same intercepts). The corresponding augmented matrix for this -6 2 -8 system is = G. 3 -1 4 Replace the second row by itself +  12 times the first row. This gives -6 2 -8 G. We could complete the Gauss–Jordan algorithm by dividing = 0 0 0 1 4 the first row by –6, giving the reduced row echelon form of > 1 - 3 3 H. 0 0 0 The variables in this example are x and y. There is one leading 1, corresponding to the x variable (as the coefficients of x were in the first column), and so we describe variable x as basic or leading and the other variable, y, as free. The second row now tells us that 0y = 0, which is true for any value of y, so we let y = k where k ∈ R is an arbitrary constant. 1 4 Then, from the first row of the reduced matrix, we have x - 3 y = 3 , 1 4 k+4 and, since y = k , x = 3 k + 3 = 3 . Thus, we can write the solution (in parametric form) as $` k +3 4 , kj: k d R. .

In general, the solution to consistent systems such as the one just considered, but also to more complicated cases where there are both leading (corresponding to leading 1s) and free variables, is written by assigning arbitrary constants, or parameters, to the free variables, and then writing the leading variables in terms of these arbitrary constants. Most CAS have a function which automatically reduces an augmented matrix to reduced row echelon form, from which the solution can be determined. This is fine unless there is an arbitrary constant in the augmented matrix itself (for example, arising from one of the linear equations in the system involving an arbitrary constant for one of the coefficients or its

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chapter 3 Solving systems of simultaneous linear equations

constant term). Then it is necessary to be wary that a division by a function of the constant may have taken place, and that this operation will only be valid if the function of the constant is non-zero. The resulting reduced row echelon matrix may have no trace of the constant. Some CAS have a ‘fraction free Gaussian-elimination’ function which effectively only gives the row echelon matrix, and this can be used to investigate what happens for such systems.

S t u d e n t ac t i v i t y 3 . 3 For the following systems of equations, enter the augmented matrix into a CAS, or other suitable technology, and use this to obtain the reduced row echelon form. Hence solve the following systems of simultaneous linear equations. Z _ ] x + 2y - z = 2b a [ x + 4y - 3z = 3` ]2x + 5y - 3z = 1b \ a Z _ x + y + z = 1b ] b [ x - y + 3z = 5` ] 3x - 2y + z = - 2b \ a Z _ ] 2x + 2y - z = 5b c [- 2x + y + z = 7` ]- 4x + y + 2z = 10b \ a

S y s t e m s o f s i m u l t a n e o u s l i n e ar e q u a t i o n s i n v ari o u s c o n t e x t s Many different contexts give rise to systems of simultaneous linear equations in several variables, even when the relations involved may themselves be nonlinear. Substitution of values of the variables in such contexts often results in a system of simultaneous linear equations relating coefficients and/or arbitrary constants.

59

MathsWorks for Teachers Matrices Example 3.10

A circle has an equation of the form x 2 + y2 + ax + by + c = 0, where a, b and c ∈ R. This circle passes through the points (–2, 3), (6, 3) and (2, 7) in the cartesian plane. Find the values of a, b and c. Solution

Since (–2, 3) lies on the circle, substitution of these values into the equation for the circle gives (–2)2 + 32 – 2a + 3b + c = 0. This simplifies to the linear equation:

–2a + 3b + c = –13 Similarly, as (6, 3) also lies on the circle we have 62 + 32 + 6a + 3b + c = 0, which simplifies to the linear equation:

6a + 3b + c = –45 For the point (2, 7), which also lies on the circle, we have 22 + 72 + 2a + 7b + c = 0, and hence:

2a + 7b + c = –53 We thus have a system of three simultaneous linear equations in a, b and c: Z _ ]- 2a + 3b + c =- 13b [ 6a + 3b + c =- 45` ] b \ 2a + 7b + c =- 53a The augmented matrix form is: R V S- 2 3 1 - 13 W S 6 3 1 - 45W SS 2 7 1 - 53 WW T X and using by hand or CAS manipulation to bring this to reduced row echelon form yields: R V S1 0 0 - 4 W S0 1 0 - 6 W SS0 0 1 - 3 WW T X The solution can be read directly from this: a = – 4, b = – 6 and c = – 3, and so the equation of the circle is x2 + y2 – 4x – 6y – 3 = 0. Completing the square, on both x and y, results in the alternative equation of the form (x – 2)2 + (y – 3)2 = 16 for the relation. The graph of this relation is a circle with centre (2, 3) and radius 4, as shown in Figure 3.9.

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chapter 3 Solving systems of simultaneous linear equations

y 8 6 4 2

–6

–4

–2

O

2

4

6

8

x

–2 –4 –6 Figure 3.9: Graph of the relation x2 + y2 – 4x – 6y – 3 = 0 or (x – 2) 2 + (y – 3) 2 = 16

Example 3.11

Three Toyotas, two Fords and four Holdens can be rented for $212 per day. Alternatively, two Toyotas, four Fords and three Holdens can be rented for $214 per day, or four Toyotas, three Fords and two Holdens could be rented for $204 per day. Assuming that the rate for renting any type of car is fixed by the make, find the rental rates for each type of car per day. Solution

Let a, b and c be the respective costs of renting a Toyota, a Ford and a Holden per day. Then we have three simultaneous linear equations in the three unknowns a, b and c.



3a + 2b + 4c = 212 2a + 4b + 3c = 214 4a + 3b + 2c = 204

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MathsWorks for Teachers Matrices

The augmented matrix corresponding to this system is: R V S 3 2 4 212 W S 2 4 3 214 W SS 4 3 2 204 WW T X and using by hand or CAS manipulation to return the reduced row echelon form yields: R V S1 0 0 20 W S0 1 0 24 W SS0 0 1 26 WW T X Hence the rental rates are $20 per day for a Toyota, $24 per day for a Ford and $26 per day for a Holden.

Example 3.12

A girl finds $5.20 in coins: 50 cent coins, 20 cent coins and 10 cent coins. She finds 21 coins in total. How many coins of each type must she have? Solution

Suppose she has a 50 cent coins, b 20 cent coins and c 10 cent coins. Then

50a + 20b + 10c = 520 She has 21 coins in total, so:



a + b + c = 21 This gives us two simultaneous linear equations in three unknowns. We write the augmented matrix corresponding to the system: =

50 20 10 520 G 1 1 1 21

and find its reduced row echelon form: R V S1 0 - 1 10 W 3 3W S S0 1 4 53 W S 3 3W T X Generally, there would be infinitely many possible solutions to these equations, but we require non-negative integers as solutions. The leading variables correspond to the columns with leading 1s, so are a and b.

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chapter 3 Solving systems of simultaneous linear equations

The free variable is c, and it can take integer values between 0 and 21. We can write a and b in terms of c, from the reduced row echelon matrix above, as 10 c 10 + c a= 3 +3= 3 53 4c 53 - 4c b= 3 - 3 = 3



To find the possible integer solutions, we need to consider integer values of c between 0 and 21, and determine when 10 + c and 53 – 4c are both divisible by 3. This could easily be done by technology, forming a 22 × 3 matrix, with 10 + c 53 - 4c the first column containing c, the second 3 and the third 3 .

c 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

10 + c 3 3.3 3.7 4.0 4.3 4.7 5.0 5.3 5.7 6.0 6.3 6.7 7.0 7.3 7.7 8.0 8.3 8.7 9.0 9.3 9.7 10.0 10.3

53 - 4c 3 17.7 16.3 15.0 13.7 12.3 11.0 9.7 8.3 7.0 5.7 4.3 3.0 1.7 0.3 –1.0 –2.3 –3.7 –5.0 –6.3 –7.7 –9.0 –10.3

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Thus we see that there are several possibilities: {a = 4, b = 15, c = 2} or {a = 5, b = 11, c = 5} or {a = 6, b = 7, c = 8} or {a = 7, b = 3, c = 11} Example 3.13

The scores of three players in a tournament have been lost. The only information available is the total of the scores for players 1 and 2, the total for players 2 and 3, and the total for players 3 and 1. Show that the original scores can be recovered. Solution

Let x, y and z be the scores for players 1, 2 and 3 respectively, and a, b and c the totals for players 1 and 2, 2 and 3, and 3 and 1 respectively. Then x+y = a y+z = b z+x = c

is a system of three simultaneous linear equations in three unknowns x, y and z. The augmented matrix is: R V S1 1 0 a W S0 1 1 b W SS W 1 0 1 cW T X Its corresponding reduced row echelon form is R V S1 0 0 a - b + c W 2 S W a+b-c W S S0 1 0 W 2 S -a + b + c W SS0 0 1 WW 2 T X a-b+c a+b-c So the original scores are x = , and y = 2 2 z=

64

-a + b + c . 2

chapter 3 Solving systems of simultaneous linear equations Example 3.14

Find the rule for the family of parabolas which pass through the points (1, 2) and (3, 4). Solution

Let the rule for the family of parabolas be y = ax 2 + bx + c, where a is non-zero. Since (1, 2) lies on any member of this family of curves:

a+b+c = 2 Similarly, since (3, 4) lies on any member of the family of curves:

9a + 3b + c = 4 Hence we have a system of two equations in three unknowns a, b and c. We first write the augmented matrix:

=

1 1 1 2 G 9 3 1 4

and use CAS to reduce it to its reduced row echelon form:

>



1 0 - 13 - 13 0 1

4 3

7 3

H

Now a and b are the leading variables, and c is a free variable, so we c 1 let c = k where k ∈ R. Then the first row gives a - 3 =- 3 , and so a=

4 7 c-1 k-1 3 = 3 . The second row gives b + 3 c = 3 and so

7 - 4c 7 - 4 k 3 = 3 . Hence the rule for the family of parabolas is:

b=



y=

k - 1 2 7 - 4k 3 x + 3 x + k , k ∈ R

We will graph a few of these curves. If k = 1, then a = 0 and we obtain the straight line with equation y = x + 1, which passes through the two points. If k = 4, then we obtain the parabola with equation y = x 2 – 3x + 4. If k = – 5, then we obtain the parabola y = –2x 2 + 9x – 5. The graphs of these curves are shown in Figure 3.10.

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MathsWorks for Teachers Matrices

y 6 g

5

h

4

(3, 4)

3 2 1

O

–1

f (1, 2)

1

2

3

4

x

–1 –2 Figure 3.10: Graph of f(x) = –2x2 + 9x – 5, g(x) = x2 – 3x + 4 and h (x) = x + 1

In the above example, we found a and b in terms of c. We could have used the same procedure to find, say, b and c in terms of a. All we would need to do is to have the columns in the matrix corresponding to b and c come before the column corresponding to a. In this case, the augmented matrix would be 1 0 4 1 1 1 1 2 G, and the reduced row echelon form would be = = G. Now we 0 1 -3 1 3 1 9 4 let a = r, where r ∈ R. Then from the first row we have b = 1 – 4r and from the second row we have c = 1 + 3r, giving the form of the family of parabolas as y = rx 2 + (1 – 4r)x + 1 + 3r where r ∈ R. Students may find it of interest to explore what values of k and r are required to produce a collection of given quadratic functions. Example 3.15

Find the rule for the family of cubic polynomials which passes through the points (1, 0) and (–1, 0), with slope –4 when x = 1. Solution

Let f(x) = ax3 + bx2 + cx + d be the rule of a cubic polynomial function, with a, b, c and d the unknown coefficients. Since (1, 0) lies on the curve:

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a+b+c+d = 0

chapter 3 Solving systems of simultaneous linear equations

Similarly, since (–1, 0) lies on the curve:

-a + b - c + d = 0

Now f ′(x) = 3ax 2 + 2bx + c, and the slope at x = 1 is –4, so 3a + 2b + c = –4. We now have a system of three simultaneous linear equations in four unknowns: Z _ ] a + b + c + d = 0b [- a + b - c + d = 0` ] 3a + 2b + c = - 4b \ a We can write down the augmented matrix corresponding to this system: R V S 1 1 1 1 0 W S- 1 1 - 1 1 0 W S W S 3 2 1 0 - 4W T X The reduced row echelon form of this is: R V S1 0 0 - 1 - 2 W S0 1 0 1 0 W SS W 0 0 1 1 2 W T X There are three leading variables, a, b and c, and one free variable, d. Let d = k where k ∈ R, and express the leading variables in terms of k. The first row of the reduced matrix tells us that a – d = –2, and so a = –2 + k. The second row tells us that b + d = 0, and so b = –k and the third row tells us that c + d = 2, so c = 2 – k. Thus, the rule for the family of polynomials is: f: R → R, where f(x) = (k – 2)x3 – kx2 +(2 – k)x + k, and k ∈ R. If k = 2, the function will be the quadratic with rule f(x) = –2x 2 + 2. We can check that this is the general form by drawing the graphs of some members of this family: If k = 0, f(x) = –2x3 + 2x. If k = 3, f(x) = x3 – 3x 2 – x + 3. If k = –1, f(x) = –3x3 + x 2 + 3x – 1. The graphs of these, and for k = 2, are shown in Figure 3.11 on the following page.

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MathsWorks for Teachers Matrices y 4 y = –2x + 2x

p

3

3 y = x3 –3x2 – x + 3 f

2 1

–4

–3

–2

y = –3x + x + 3x – 1 3

–1 g

2

y = –2x2 + 2

h

O –1

1

2

3

4

x

–2 –3 –4

Figure 3.11: Graphs of f(x) for k = –1, 0, 2 and 3

Example 3.16

Find the rule for the family of quartic polynomials (polynomials of degree 4) that pass through the points (1, 2) and (–2, –1), and have slope 5 at x = 1. Solution

Let f(x) = ax4 + bx3 + cx 2 + dx + e be the rule for the family of quartic polynomials. Since they pass through (1, 2), f(1) = 2, so:

a + b + c + d + e = 2 Since they pass through (–2, –1), f(–2) = –1, so:



16a – 8b + 4c – 2d + e = –1 Now f ′(x) = 4ax3 + 3bx 2 + 2cx + d. Since we must have f ′(1) = 5:



4a + 3b + 2c + d = 5 We now have a system of three simultaneous linear equations in the five unknowns a, b, c, d and e: Z _ ] a + b + c + d + e = 2b [ 16a - 8b + 4c - 2d + e = - 1` ] 4a + 3b + 2c + d = 5b \ a

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chapter 3 Solving systems of simultaneous linear equations

The augmented matrix for this system is: R V S1 1 1 1 1 2 W S16 - 8 4 - 2 1 - 1W SS W 4 3 2 1 0 5 W T X which has reduced row echelon form: R V S1 0 0 - 1 - 3 1 W 4 12 W 2 S 1 5W S S0 1 0 0 - 2 6 W S 3 9 13 W SS0 0 1 2 4 12 WW T X So the leading variables are a, b and c, and the free variables d and e. Let d = s and e = t where s, t ∈ R. Our solution will now be given in terms of two parameters. Then the first row of the reduced echelon 1 3 1 matrix corresponds to the equation a - 2 d - 4 e = 12 , hence 1 3 1 a = 2 s + 4 t + 12 . The second row of the reduced echelon matrix corresponds to the 1 5 1 5 equation b - 2 e = 6 , hence b = 2 t + 6 . The third row of the reduced echelon matrix corresponds to the 3 9 13 3 9 13 equation c + 2 d + 4 e = 12 , hence c =- 2 s - 4 t + 12 .

The solution set is 'b 1 s + 3 t + 1 , 1 t + 5 , - 3 s - 9 t + 13 , s, t l: s, t d R1 4 4 2 12 2 6 2 12

and the family of functions has the form: 1 3 1 1 5 3 9 13 f]xg = b 2 s + 4 t + 12 l x 4 + b 2 t + 6 l x 3 + b- 2 s - 4 t + 12 l x2 + sx + t where s, t d R . As a solution, this looks fairly formidable, so it’s a useful strategy to plot a few of its members. • If s = 0 and t = 0 , then we have the function 1 5 13 f]xg = 12 x 4 + 6 x 3 + 12 x2 . • If s = 1 and t = 0 , then we have the function 7 5 5 f]xg = 12 x 4 + 6 x 3 - 12 x2 + x .

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MathsWorks for Teachers Matrices

• If s = 0 and t = 1 , then we have the function 5 4 7 f ] xg = 6 x 4 + 3 x 3 - 6 x 2 + 1 . • If s = 1 and t = 1 , then we have the function 4 4 8 f ] xg = 3 x 4 + 3 x 3 - 3 x 2 + x + 1 . • If s =- 1 and t =- 1 , then we have the function 7 1 29 f ] xg = - 6 x 4 + 3 x 3 + 6 x 2 - x - 1 . These members are shown in Figure 3.12 below. y 5 4 3 2 1 –4

–3

–2

–1

O –1

1

2

3

4

x

–2 –3 –4 –5 Figure 3.12: Parts of the graphs of some members of the family of functions for various values of s and t

S t u d e n t ac t i v i t y 3 . 4 a

b c d

The scores of four players in a tournament have been lost. The only information available is the total of the scores for players 1 and 2, the total for players 2 and 3, the total for players 3 and 4 and the total for players 4 and 1. Can the original scores be recovered? Find the equation of the cubic polynomial which passes through the points (1, 0) and (–1, 0), with slope –4 when x = 1 and slope 12 when x = –1. Find the rule for the family of quartic polynomials (polynomials of degree 4) that pass through the points (1, 2), (–2, –1) and (2, 0), and have slope 5 at x = 1. Find the rule for the quartic polynomial (polynomial of degree 4) that passes through the points (1, 2), (–2, –1), (2, 0) and (–1, 5), and has slope 5 at x = 1.

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chapter 3 Solving systems of simultaneous linear equations

SUMM A R Y

• A linear system of m simultaneous equations in n variables x1, x2, f, xn is a set of m equations of the form a11 x1 + a12 x2 + f a1n xn = b1 a21 x1 + a22 x2 + f a2n xn = b2 am1 x1 + am2 x2 + f amn xn = bm • The numbers a11, a12, f, a1n, f, amn are the coefficients of the system, and b1, b2, f, bm are the constant terms. • A system of simultaneous linear equations is said to be consistent if it has either a unique solution or infinitely many solutions, and inconsistent if it does not have a solution. • For a system of two simultaneous linear equations in two variables ]m = n = 2g: – there is a unique solution which corresponds to the point of intersection of the graphs of the corresponding straight lines (different gradient) in the cartesian plane, R2 or – there are infinitely many solutions which correspond to the infinite set of points that comprise the superimposition of the graphs of the same straight line (same gradients and same axis intercepts) specified by two equivalent linear relations, in the cartesian plane, R2 or – there are no solutions, and the graphs of the corresponding straight lines are parallel (same gradient) but distinct (different axis intercepts) straight lines in the cartesian plane, R2. • For a system of three simultaneous linear equations in three variables ]m = n = 3g: – there is a unique solution which corresponds to the point of intersection of the graphs of the corresponding planes in threedimensional space, R3 or – there are infinitely many solutions which correspond to the graphs of three planes aligned like ‘pages’ (which may include a superimposed page or pages) in a book along a common spine, in three-dimensional space, R3, and where at least two of these ‘pages’ are distinct, their intersection points form a straight line in R3

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SUMM A R Y (Cont.)

or – there are no solutions, and the graphs of the three planes are all parallel but distinct; or one pair is parallel (and distinct) and the other oblique to this pair; or they are configured like a triangular prism. • The system of simultaneous linear equations can be written in matrix form AX = B, where R V R V S a11 a12 f a1n W S x1 W S a21 a22 f a2n W S x2 W A=S W is the m × n coefficient matrix, X = S W the h h W S h ShW Sam1 am2 f amn W S xn W T X T X R V S b1 W Sb W n × 1 column matrix (vector) of variables, and B =  S 2 W the m × 1 Sh W column matrix (vector) of constant terms. Sbm W T X • If A is an invertible (non-singular) square matrix (m = n) then the inverse method can be employed and X = A–1B. • The m × (n + 1) augmented matrix of the system is the following matrix: R S a11 a12 f a1n S a21 a22 f a2n S h h S h Sam1 am2 f amn T

V b1 W b2 W W h W bm W X

• The system of simultaneous linear equations of the form AX = O is said to be homogeneous and is always consistent, with X = O (the relevant zero vector) a solution. • To solve such systems of equations using the Gauss–Jordan method, there are three steps. Step 1: Write the augmented matrix for the system of equations. Step 2: Enter the augmented matrix into CAS, or other suitable technology, and obtain the reduced row echelon form of the matrix (using exact arithmetic where possible).

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chapter 3 Solving systems of simultaneous linear equations

SUMM A R Y (Cont.)

Step 3: Interpret the resulting reduced row echelon matrix, as follows: Case 1: If the number of leading 1s is equal to the number of variables, and the last leading 1 is not in the rightmost column, then there is a unique solution which can be written down directly from the matrix. Case 2: If the number of leading 1s is less than the number of variables, and the last leading 1 is not in the rightmost column, then there will be an infinite number of solutions. The solutions can be written by assigning an arbitrary constant to each of the free variables (those not corresponding to leading 1s), and writing the leading variables in terms of these constants. Case 3: If the last non-zero row of the reduced row echelon matrix has a leading 1 in the rightmost column, then the system of equations is inconsistent (that is, has no solution). • Non-linear functions and relations can be used to generate a system of simultaneous linear equations where substitution of some values for the variables leads to such a system expressed in terms of coefficients used to specify the particular functions and/or relations involved.

References Anton, H & Rorres, C 2005, Elementary linear algebra (applications version), 9th edn, John Wiley and Sons, New York. Cirrito, F (ed.) 1999, Mathematics higher level (core), 2nd edn, IBID Press, Victoria. Hill, RO Jr 1996, Elementary linear algebra with applications, 3rd edn, Saunders College, Philadelphia. Lipschutz, S & Lipson, M 2000, Schaum’s outline of linear algebra, 3rd edn, McGraw-Hill, New York. Nicholson, KW 2001, Elementary linear algebra, 1st edn, McGraw-Hill Ryerson, Whitby, ON. Nicholson, KW 2003, Linear algebra with applications, 4th edn, McGraw-Hill Ryerson, Whitby, ON. Poole, D 2006, Linear algebra: A modern introduction, 2nd edn, Thomson Brooks/ Cole, California. Wheal, M 2003, Matrices: Mathematical models for organising and manipulating information, 2nd edn, Australian Association of Mathematics Teachers, Adelaide.

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Websites http://en.wikipedia.org/wiki/Gaussian_elimination – Wikipedia This site provides a comprehensive discussion with links to other resources and references. http://mathworld.wolfram.com/GaussianElimination.html – Wolfram Research This site is the online mathematical encyclopaedia from the developers of the CAS Mathematica. It provides a concise but comprehensive mathematical overview and includes links to related topics and a good list of other references. http://www.sosmath.com/matrix/system1/system1.html – SOS Mathematics This site provides an accessible discussion with worked examples using Gaussian elimination for some simple cases of systems of simultaneous linear equations. http://aleph0.clarku.edu/~djoyce/ma105/simultaneous.html – Department of Mathematics and Computer Science, Clark University. This site includes a first principles discussion of a practical problem based on ancient Chinese methods. http://www.jgsee.kmutt.ac.th/exell/PracMath/SimLinEq.html – Practical Mathematics. This site covers straightforward examples for 2 × 2 and 3 × 3 systems, with a collection of related exercises. http://mathforum.org/linear/choosing.texts/ – Drexel University This site provides information on selected linear algebra texts, including those that are technology based. http://www.sosmath.com/matrix/matrix.html – SOS Mathematics This site has some notes with examples on solving systems of linear equations, and is at quite a simple level.

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C ha p t e r

4

Tra n s f o r m a t i o n s o f t h e car t e s ia n p l a n e A transformation on the cartesian plane, R × R, or R2 as it is also commonly designated, is a correspondence or mapping from the set of points in the plane to a set of points in the plane. That is, for every (original) point in the plane before the transformation is applied, there is a corresponding unique point, the image, in the plane after the transformation is applied. In senior secondary mathematics curricula, particular importance is assigned to the study of the effects of transformations on certain subsets of the plane—those that correspond to the graphs of functions and other relations. Possibly the first case of analysis related to transformations of (graphs of) functions for middle school secondary students is the graph of the function with rule g(x) = a(x + b)2 + c, derived from the graph of the function with rule f(x) = x 2 by a sequence of transformations involving a dilation from the y-axis, possibly a reflection in the x-axis (depending on the sign of a), a translation parallel to the x-axis and a translation parallel to the y-axis, although a good case could be made for considering graphs of linear functions of the form y = mx + c as a similar sequence of transformations of the graph of y = x. The first two transformations mentioned above—dilation and reflection— are examples of what are commonly called linear transformations, while all three of these transformations are examples of what are called affine transformations (linear transformations and also translations). Some care will need to be taken for students to become clear that there are two types of ‘function’ involved in this context: the function of a single real variable whose graph corresponds to a particular type of subset of the cartesian plane {(x, y) where y = f(x) and x ∈ domain(f)}, and the function which is a transformation of the plane that maps an ordered pair (that is, a point) to another unique ordered pair.

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Matrices are well suited to the analysis of linear transformations, and provide a convenient notation for distinguishing between these two senses of function. Indeed, linear transformations can be used to provide a strong motivation for the definition of matrix multiplication, as applied to 2 × 2 matrices. The application of matrices to the analysis of linear transformations involves the solution of systems of simultaneous linear equations and matrix inverses.

Li n e ar t ra n s f o r m a t i o n s A linear transformation is a function T:R2 → R2, T(u) = w, where u and w are ordered pairs corresponding to points in the plane, which satisfies the following properties or axioms, called the linearity axioms: T(u + v) = T(u) + T(v) for all u and v ∈ R2 T(kv) = kT(v) for all v ∈ R2 and all scalars k ∈ R More generally a linear transformation is defined as a function from one vector space (see Chapter 2, page 20) to another that satisfies the linearity axioms above. In this text we will consider only the restricted case of R2, where the underlying vector space is that of coordinate vectors in the cartesian plane. It is important that teachers clarify for students the nature of the cartesian plane as R × R, or R2. Students may or may not have come across the notion of the cartesian product of two sets X and Y, where X × Y = {(x, y): x ∈ X and y ∈ Y}. Even if they are familiar with this notion, for example, from listing the event space for two events with a finite discrete set of possible outcomes, they may not transfer this conceptually to the case of uncountable continuous sets, or at least require reminding of its application in this context. Indeed, their own practical experience is much more likely to have them familiar with a well known subset of R2, that is, the set of all integer (whole number) valued grid points, part of which is shown in Figure 4.1, and which constitute Z × Z = Z2, where Z is the set of integers. In the case of R2 = {(x, y): x, y ∈ R}, where X = Y = R, the corresponding cartesian product can be regarded as the set of all points in the cartesian plane or the set of all position vectors for these points with respect to the origin of the cartesian plane. Clearly there is a one-to-one correspondence between points and position vectors with respect to the origin, and matrix notation is quite useful in this context.

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chapter 4 Transformations of the cartesian plane y 8 6 4 2 –8

–6

–4

–2

O

x 2

4

6

8

–2 –4 –6 –8 Figure 4.1: Intersecting lines indicating a subset of grid points of Z × Z = Z   2, where Z is the set of integers

x Since any u = (x, y) ∈ R2 can be written as a column matrix u = = G, we y 1 0 can write u = x = G + y = G; that is, we can write u as a linear combination of 0 1 1 0 = G and = G. 0 1 By the linearity properties, to determine the image of u under T we simply 1 0 need to determine the image of = G and = G under T. The set of vectors 0 1 1 0 {[1, 0], [0, 1]} or )= G, = G3 is said to be a basis of the vector space they 0 1 generate, in this case the set of all coordinate vectors in the cartesian plane, as x any coordinate vector ^x, yh = = G is a linear combination of these two vectors. y This simple but powerful concept underpins much of the work relating to transformations of the plane, so it is useful to take some time to ensure that students have a sound grasp of it. In work on vector representations in the 1 0 plane, )= G, = G3 corresponds to the unit vectors commonly denoted {i, j} 0 1 (see Evans, 2006, Chapter 8).

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a 1 0 b Let T e= Go = = G and T e= Go = = G for some real numbers a, b, c, and d. c 0 1 d x 1 0 1 0 Then T e= Go = T e x = G + y = Go = xT e= Go + yT e= Go y 0 1 0 1 ax + by a b a b x = x= G + y= G = = G= G G== cx + dy c d c d y Such a linear transformation can be accomplished by a matrix multiplication, with the matrix determined by the image of the two points 1 0 a b = G and = G. Conversely, any 2 × 2 matrix = G can be considered a linear 0 1 c d a 1 0 b transformation that transforms = G to the point = G and = G to the point = G. c 0 1 d Any linear transformation T:R2 → R2 can be written in the form T(u) = Au, where A is a 2 × 2 matrix. If the transformation T is applied to a region with area s, then the area of the transformed region is equal to the product of s and the absolute value of the determinant of A, that is |det(A)| × s. Moreover, A can be uniquely determined if the images of two points u and v a b are known, where u ≠ kv for some non-zero k ∈ R. If A = = G is the matrix c d for T, then: a 1 0 a b 1 a b 0 b T e= Go = = G = G = = G and T e= Go = = G= G = = G c 0 1 c d 0 c d 1 d



Example 4.1

0 1 1 0 Let T be the transformation with matrix = G. Then T e= Go = = G and 1 0 0 1 0 1 T e= Go = = G. 1 0 a Find the image of the point (2, 4). b Find the image of the point (x, y). Solution

a To find the image of the point (2, 4), we simply find the matrix product:

78

=

0 1 2 4 G= G = = G 1 0 4 2

chapter 4 Transformations of the cartesian plane

So the point (2, 4) is transformed or mapped to the point (4, 2). y 0 1 x b Since = G = G = = G, any point (x, y) is mapped to the point (y, x). x 1 0 y Geometrically, this transformation corresponds to a reflection in the line y = x, and can be used to determine inverse relations. 1 0 If the points whose images are known are not = G and = G, then pairs of 0 1 simultaneous linear equations or inverse matrices are used to find the matrix for the transformation. Example 4.2

If T is a linear transformation that transforms (1, 2) to (–1, 1) and (3, 1) to (0, 1), find the matrix for this transformation. Solution

Let =

a b G be the matrix corresponding to this linear transformation. c d

Then =

-1 0 a b 1 a b 3 G = G = = G and = G = G = = G. 1 1 c d 2 c d 1

a + 2b =- 1 3a + b = 0 and . c + 2d = 1 3c + d = 1 Combining the four equations, we have two equations involving a and b, and two equations involving c and d. For integer values of a, b, c and d, these can usually be readily, if somewhat tediously, solved by hand. a + 2b = - 1 c + 2d = 1 1 3 ) 3 has solution a = 5 , b =- 5 , and ) 3 has 3a + b = 0 3c + d = 1 R V S1 - 3W 1 2 solution c = 5 , d = 5 , hence the required matrix is S 15 25 W. S W S5 5 W T X That is,

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Alternatively, = and written as =

=

-1 a b 1 a b 3 0 G = G = > H and = G = G = = G can be combined 1 c d 2 c d 1 1

-1 0 a b 1 3 H and then G= G = > c d 2 1 1 1

- 1 0 1 3 -1 a b H= G G=> c d 1 1 2 1 -1

-1 0 5 => H> 1 1 25 1 -3 5 5 2 5 5

= >1

3 5 -1 5

H

H

Example 4.3

-1 2 The matrix A = > H transforms the point P(x, y) onto the point 3 -8 Q(1, –7). Find the coordinates of the point P. Solution

-1 2 x 1 We have = G = G = = G. 3 -8 y -7 -4 -1 1 - 1 2 -1 1 x 3 Hence = G = = G = G = > 3 1 H= G = = G - 2 - 2 -7 3 -8 -7 y 2 So P has coordinates (3, 2).

S t u d e n t ac t i v i t y 4 . 1 a

Show that any linear transformation maps the origin to the origin.

1 3 -1 -3 b Explain why a linear transformation T with T f= Gp = = G and T e= Go = = G is -1 -1 1 1 c

not uniquely determined. Find at least two linear transformations that satisfy the above conditions. Find the points that are mapped to the points (1, 0) and (0, 1) by the linear 4 3 transformation with matrix = G. 5 4

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chapter 4 Transformations of the cartesian plane

Li n e ar t ra n s f o r m a t i o n o f a s t rai g h t l i n e Although a linear transformation acts upon individual points or position vectors, we usually want to see how a set of points corresponding to a subset of the plane of interest to us is transformed, particularly curves and figures. The transformation T:R2→ R2 with the rule T(x, y) = (x + y, x – y) can be written in matrix form, with (x, y) as an original point and (x1, y1) as an image point: x1 1 1 x x 1 1 1 x1 > H== G= G , = G = 2= G> H y1 1 -1 y 1 - 1 y1 y



1 1 Hence x = 2 ^x1 + y1h and y = 2 ^x1 - y1h. Under this transformation, the image of the graph of the relation with the equation ax + by + c = 0 (which is a straight line) is the graph of the relation with the equation 1] 1] a b g g 2 ^x1 + y1h + 2 ^x1 - y1h + c = 0 , or 2 a + b x1 + 2 a - b y1 + c = 0 (which is also a straight line). In particular, the straight line with equation 2x + 3y - 6 = 0 is transformed onto the straight line with equation 5 1 2 x - 2 y - 6 = 0 , as shown for part of the original line (that is a line segment subset of the original line) and the corresponding image points in Figure 4.2. y +

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

–4

4

2

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

2x + 3y – 6 = 0 5x – y – 6 = 12

O

–2

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

–2

–4

2

4

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

Figure 4.2: Graph of 2x + 3y – 6 = 0, 0 < x < 3, and its image under T(x, y) = (x + y, x – y)

81

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MathsWorks for Teachers Matrices

An equation for any straight line can easily be written in parametric form. Although not commonly used for straight lines in the cartesian plane (unless as a simple application of vector kinematics), the parametric form of an equation for a straight line is simply a vector equation for the line. It gives the position vector of each point on the line with respect to the origin. In this sense the coordinates of a point in the plane correspond to its position vector relative to the origin (0, 0). Parametric forms are very useful in computer graphic applications. The following discussion should be developed through an exposition that connects the graphical picture with the conceptual and symbolic argument. Consider the straight line that passes through the two distinct points P, with position vector p = (x1, y1), and Q, with position vector q = (x 2, y2). A direction vector for this line from P to Q is d = q – p = (x 2 – x1, y2 – y1), and the position vector r of any point on the line is given by r = p + td, t ∈ R. That is, any point on the line that passes through P and Q must be some distance (a scalar multiple of the length of the directed line segment PQ ) along this line from the point P. This is illustrated in Figure 4.3. If we restrict t to the interval [0, 1], then we have exactly the directed line segment from P to Q. As d = q – p, another way of writing this position vector is r = p + t(q – p) = (1 – t)p + t q. When t = 0, r = p, and when t = 1, r = q, the endpoints of the line segment. If 0 ≤ t ≤ 1, then the vector r is clearly the position vector of some point on the directed line segment PQ . Q

d

P

R

r p

q

O Figure 4.3: Vector representation of a line through two distinct points in the plane

For example, consider the line passing through the points P(1, 2) and Q(–3, 4). Then the position vector r of any point R on the line through P and Q can be written r = (1, 2) + t (–4, 2), where PQ = (–4, 2).

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chapter 4 Transformations of the cartesian plane

It is natural for students to inquire how this relates to the more common representation of a straight line by the rule y = mx + c, where c is interpreted as the y-axis intercept and m represents the slope of the line (the ratio of relative difference between the y values of two points with respect to their corresponding difference in x values). A simple parameterisation of y = mx + c is to let x = t, where t ∈ R, then y = mt + c and the corresponding vector parametric form is (t, mt + c), t ∈ R. This simple parameterisation can x 0 1 also be written in matrix form as = G = = G + t = G. y c m In vector terms, the position vector of any point S on the line is the sum of 0 the vector = G, the position vector of the point (0, c), the y-axis intercept, and c 1 a scalar multiple of the vector = G, which gives the direction of the line. Note m 1 that the vector = G has a horizontal component of 1 unit and a vertical m component of m units, as shown in Figure 4.4.

S y

(0, c)

m 1

O

x

Figure 4.4: Vector representation of y = mx + c in the plane

Using matrices and vectors, it can easily be shown that under a linear transformation with non-singular matrix: 1 The origin is mapped onto itself. 2 The transformation is a one-to-one mapping of R2 onto R2. 3 A straight line (line segment) is mapped onto a straight line (line segment).

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4 Any pair of distinct parallel lines is mapped onto another pair of distinct parallel lines. 5 A straight line that passes through the origin is mapped onto another straight line that passes through the origin. If the matrix of the transformation is singular, then any line will be mapped to a line or point. There are several methods for determining the equation of the image of a line under a linear transformation. Some involve the use of the inverse of a 2 × 2 matrix, while others do not. The following discussion illustrates three different approaches. Teachers should encourage students to think about the mathematical strategies involved, and where various constructs arise in each case. Example 4.4

Consider the linear transformation associated with the matrix 1 2 A == G. Find the image of the line with rule y = 2x + 3 under this 4 -3 transformation. Solution Method 1

Since straight lines are transformed onto straight lines under a linear transformation, we can find the image by finding the image of any two distinct points on the original straight line. For example, the points (0, 3) and (1, 5) clearly lie on the straight line with equation y = 2x + 3.The corresponding image points are given by =

1 2 0 6 1 2 1 11 G = G = = G and = G= G = = G. 4 -3 3 -9 4 -3 5 - 11

So the image of the line y = 2x + 3 passes through the points (6, –9) and (11, –11). y -y Using the general form y - y1 = d x2 - x 1 n ^x - x1h, we get 2

2 1 ^y + 9h =- 5 ]x - 6g or y =- 5 ]2x + 33g.

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chapter 4 Transformations of the cartesian plane Method 2

Use a vector parametric form for the straight line, t 0 1 r = = G + t = G = = G, t ∈ R. 3 + 2t 3 2 Then T(r) = =

1 2 t 6 + 5t 6 5 G= G== G = = G + t = G. This 4 - 3 3 + 2t - 9 - 2t -9 -2

5 corresponds to a line through the point (6, –9) in the direction of = G, -2 2 2 that is, with slope - 5 , and so has cartesian equation y + 9 =- 5 ]x - 6g, 1 or y =- 5 ]2x + 33g, as before. Method 3

Consider an arbitrary point (x, y) on the line. This is mapped to the x1 x1 x 1 2 x point (x1, y1) where A f= Gp = > H, that is = G = G = > H. This gives y1 y1 4 -3 y y us equations for x1 and y1 in terms of x and y. What we want is to solve these for x and y in terms of x1 and y1, and substitute into the original equation to give an equation involving x1 and y1. This can be done easily by multiplying both sides of the matrix equation by A–1. 1 2 -1 1 2 x 1 2 - 1 x1 H > H= G = > H > H y1 4 -3 4 -3 y 4 -3 R V R V 3 2 W S 3x1 + 2y1 W S -1 x1 1 2 x 11 11 W x1 11 W H > H= S > H = SS = G=> 4 1 4 x y y S W 4 -3 y 1 - y1 W 1 1 S W S 11 11 W 11 T X T X 4x1 - y1 3 x + 2 y = 2 c 1 11 1 m + 3, which Hence y = 2x + 3 becomes 11 1 simplifies to 5y1 =- 2x1 - 33 or y1 =- 5 ^2x1 + 33h and so the equation 1 of the transformed line is y =- 5 ]2x + 33g, as before. x1 x This method can be implemented directly by finding = G = A- 1 f> Hp, y y 1

>

substituting into y = mx + c, and then solving for y1 in terms of x1. These computations can be readily carried out in one step using a CAS. This method

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will not work if we cannot find the inverse of the transformation matrix, that is, if the transformation matrix is singular. In this case the transformation maps the plane onto a line through the origin or onto the origin itself. Method 3 is easy to adapt to finding the image of any function, and is the one we will use in general. Example 4.5

1 1 G. 1 1 Under this transformation, what is the image of the line y = 2x + 3? Consider the linear transformation associated with the matrix A = = Solution

We cannot use Method 3 because the matrix A does not have an inverse, since det(A) = 1 – 1 = 0. We begin in a similar way, and find the image of the point with coordinates (x, y).

=

x1 x+y 1 1 x H=> H G= G = > y 1 1 y x+y 1

Since the x1- and y1-coordinates are the same, and x + y is not constant, the image of the line is y = x. In fact, one can show that the image of any line other than those of the form y = –x + c is y = x, and that the image of y = –x + c is the point (c, c), which of course is on the line y = x. This transformation corresponds to a projection onto the line y = x. Projections onto lines will not be considered in any detail, since they do not occur when transforming functions.

S t u d e n t ac t i v i t y 4 . 2 a Establish each of the properties of linear transformations (with non-singular matrices) listed above. b Find the equation of the image of the line with equation y = 5 – 3x under the linear 2 1 transformation with matrix = G. 1 3 c Find the equation of the image of the line with equation y = 5 – 3x under the linear 11 transformation with matrix = G . 11

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chapter 4 Transformations of the cartesian plane d

Find the equations of lines which are mapped to points under the linear 11 transformation with matrix = G . 11 Find the image of the unit square (that is, the region bounded by line segments joining vertices (0, 0), (0, 1), (1, 1) and (1, 0)) and the area of this region under the 2 1 transformation with matrix = G. 1 3

e

Li n e ar t ra n s f o r m a t i o n o f a c u r v e To find the image of the graph of y = f(x) or f(x, y) = c under the linear transformation with matrix A, we can proceed as for a straight line, provided A has an inverse. Consider an arbitrary point (x, y) on the curve which is the graph of the function or relation we are interested in. If this is mapped to the x1 x1 x x point (x1, y1) where A = G = > H, we can equivalently write = G = A- 1 > H, y y y y 1 1 which gives expressions for x and y. These can simply be substituted into y = f(x) or f(x, y) = c to find the equation of the image function or relation. Example 4.6

Find the image of the function y = x 2 and the relation 4x 2 + y2 = 1 under 1 2 the linear transformation represented by the matrix A = > H. 4 -3 Solution

Consider an arbitrary point (x, y) on the curve, which is mapped to the x1 1 2 x point (x1, y1) where > H = G = > H. This gives us equations for x1 and y1 4 -3 y y1 in terms of x and y. What we want is to solve these for x and y in terms of x1 and y1, and substitute into the original equation of the curve to give an equation involving x1 and y1. This can be done easily by premultiplying both sides of the matrix equation by A–1.



1 2 -1 1 2 x 1 2 - 1 x1 H > H= G = > H > H y1 4 -3 4 -3 y 4 -3 R V R V 3 2 W S 3x1 + 2y1 W S -1 x1 1 2 x 11 11 W x1 11 W H > H= S > H = SS = G=> 4 1 4 x y y y S W 4 3 y 1 1 W 1 1 S W S 11 11 W 11 T X T X >

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Hence the rule of the image of the function y = x 2 is the relation 4x1 - y1 c 3x1 + 2y1 m2 . Expanding this expression and replacing x1 by = 11 11 x and y1 by y gives 44x – 11y = 9x 2 + 12xy + 4y2. The graphs of both the original function and the image relation are shown in Figure 4.5. y 10 6 2 –4 –3 –2 –1

O

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20

x

–6 –10

Figure 4.5: Graph of the function y = x2 and its image relation under transformation by the matrix A

The rule of the image of the relation 4x 2 + y2 = 1 is the relation 4c

3x1 + 2y1 m2 c 4x1 - y1 m2 + = 1. Expanding this expression and 11 11 replacing x1 by x and y1 by y we have 52x 2 + 40xy + 17y2 = 121. The graphs of the original curve and its image are shown in Figure 4.6. y 4 3 2 1 –6

–5

–4

–3

–2

–1

O 1 –1

x 2

3

4

5

–2 –3 –4 Figure 4.6: Graph of the relation 4x2 + y2 = 1 and its image relation under transformation by matrix A

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chapter 4 Transformations of the cartesian plane S t u d e n t ac t i v i t y 4 . 3 a

Find the image of y = sin(x) under the linear transformation with matrix = 3 5 G. 1 2

b

Find the image of y = x2 under the transformation with matrix =

c

11 Find the image of y = x2 under the transformation with matrix = G . 11

3 0 G. 0 2

S t a n d ar d t y p e s o f l i n e ar t ra n s f o r m a t i o n s It is a key part of many senior secondary mathematics curricula to investigate the effects of certain standard types of transformations on the graphs of familiar functions and relations. In the following discussion, such an investigation is carried out with respect to the unit square—defined as the region bounded by and including the set of four line segments joining (0, 0) with (1, 0); (1, 0) with (1, 1); (1, 1) with (0, 1) and (0, 1) with (0, 0)—and the graphs of the functions with domain R and rules f(x) = x 2 and g(x) = sin(x) respectively. Note that if a linear transformation with matrix A maps a region of area a1 onto a region of area a2, then a2 = |det(A)| × a1. Computer algebra systems can be used to good effect to apply transformations to points, find algebraic relations corresponding to the transformation of variables, carry out computation involving compositions of transformations and their inverses, and draw graphs of original and transformed sets of points in the plane. Dilations from the axes Dilation by a factor k from the x-axis

The transformation matrix is of the form =

1 0 G for k > 0. 0 k

Example 4.7

1 0 G have on the unit 0 3 square and the graphs of the functions f and g? What effect does the transformation matrix = Solution

Under this transformation, each point (x, y) is mapped to the point (x, 3y), and so the corners (vertices) of the unit square (0, 0), (1, 0), (1, 1)

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and (0, 1) are mapped to the points (0, 0), (1, 0), (1, 3), (0, 3) respectively. The square has been stretched vertically, and the resulting rectangle has area 3 square units, as shown in Figure 4.7. y 3

(1, 3)

2 (1, 1)

1

–1

O

1

2

3

x

–1 1 0 Figure 4.7: Graph of the unit square and its image under the transformation with matrix < F 0 3 (corresponding to a dilation by factor 3 from the x-axis)

In general, the point (x, y) is mapped to the point (x1, y1), where x1 x 1 0 x y G = G = = G = > H, and so x = x1 and y = 31 . = y1 0 3 y 3y The effect on the graph of y = f(x) is given by substituting for x and y y in y = x 2, which results in 31 = x12. The corresponding rule for the transformed function is y = f 1(x) = 3x 2. Part of the graph of the original function and its image function is shown in Figure 4.8. y 7 6 5 4 3 2 1 –4

–3

–2

–1

O –1

1

2

3

4

Figure 4.8: Graph of part of f(x) = x2 and its image under the transformation with matrix <

90

x 1 0 F 0 3

chapter 4 Transformations of the cartesian plane

The effect on the graph of y = g(x) is given by substituting for x and y y in y = sin ]xg, which results in 31 = sin (x1) or y1 = 3 sin ^x1h. The rule for the transformed function is g1 ]xg = 3 sin ]xg. The graphs of g and g 1 are shown in Figure 4.9. y 4 3 2 1 –4

–3

–2

–1

O –1

1

2

3

4

x

–2 –3 –4 Figure 4.9: Graph of part of g(x) = sin(x) and its image under the transformation with matrix <

1 0 F 0 3

In general, dilation by factor k from the x-axis results in y = f(x) being transformed to y = kf(x). Dilation by a factor k from the y-axis

The transformation matrix is of the form =

k 0 G for k > 0. 0 1

Example 4.8

3 0 What effect does the transformation with matrix = G have on the unit 0 1 square and the graphs of the functions f and g? Solution

3x 3 0 x G = G = = G, so each point (x, y) is mapped 0 1 y y to the point (3x, y), and so the vertices of the unit square (0, 0), (1, 0), (1, 1) and (0, 1) are mapped to the points (0, 0), (3, 0), (3, 1), (0, 1) respectively. The square has been stretched horizontally, and the area of the resulting rectangle is 3 square units, as shown in Figure 4.10. Under the transformation, =

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y (1, 1)

1

O

–1

(3, 1)

1

2

3

x

4

–1 Figure 4.10: Graph of the unit square and its image under the transformation with matrix <

3 0 F 0 1

In general, the effect on any curve or region is described by x1 3x 3 0 x x G = G = = G = > H, and so x = 31 and y = y1. = y1 0 1 y y The effect on the graph of y = f(x) is given by substituting for x and y x 2 x2 in y = x 2, which results in y1 = c 31 m = 91 . The corresponding rule for x2 the transformed function is y = f1 ]xg = 9 . Part of the graph of the original function and its image function are shown in Figure 4.11. y 4 3 2 1 –10

–8

–6

–4

–2

O

2

4

6

8

10

x

–1 Figure 4.11: Graph of part of f(x) = x2 and its image under the transformation with matrix <

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3 0 F 0 1

chapter 4 Transformations of the cartesian plane

The effect on the graph of y = g(x) is given by substituting for x and x y in y = sin(x), which results in y1 = sin c 31 m. x The rule for the transformed function is g1 ]xg = sin b 3 l. The graphs of g and g 1 are shown in Figure 4.12. y 2

1

–10

–8

–6

–4

–2

O

2

4

6

8

10

–1 Figure 4.12: Graph of part of g(x) = sin(x) and its image 3 0 under the transformation with matrix < F 0 1

In general, dilation by a factor k from the y-axis results in y = f(x) being x transformed to y = f b l. k Equal dilations from both axes (scalings) These transformations are used in scale diagrams and maps. The k 0 transformation matrix is of the form = G for k > 0. This transformation is 0 k the product of a dilation from the x-axis followed by a dilation from the 1 0 k 0 k 0 k 0 1 0 y-axis, or vice versa. That is, = G== G= G== G= G. 0 k 0 1 0 1 0 k 0 k Example 4.9

3 0 G have on the unit 0 3 square and on the graphs of the functions f and g? What effect does the transformation matrix =

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Solution

3x 3 0 x G = G = = G. Each point (x, y) is mapped to 0 3 y 3y the point (3x, 3y), and so the vertices of the unit square (0, 0), (1, 0), (1, 1) and (0, 1) are mapped to the points (0, 0), (3, 0), (3, 3), (0, 3) respectively. The square has been scaled by a factor of 3 and the resultant square has an area of 9 square units, as shown in Figure 4.13. Under the transformation =

y (3, 3)

3 2 (1, 1)

1

–1

O

1

2

3

x

–1 Figure 4.13: Graph of the unit square and its image under the transformation with matrix <

Since =

3 0 F 0 3

x1 3x 3 0 x x y G = G = = G = > H, we have x = 31 and y = 31 , so the y1 0 3 y 3y

graphs of y = x 2 and y = sin(x) are transformed to the graphs of the y x y x 2 x x2 functions 31 = c 31 m and 31 = sin c 31 m or y = 3 and y = 3 sin b 3 l respectively. In general, dilation by factor k from both axes results in y = f(x) being x transformed to y = kf b l. k After a suitable range of specific examples have been explored from first principles for a variety of functions and relations, students can be led to consider the general case of a composition of dilations from both axes. This can be used to naturally and informally introduce the notion of composition of linear transformations. It also provides an example of a set of matrices for ky 0 which multiplication is commutative. Thus, given the transformation > H 0 1

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1 0 for ky > 0 (a dilation from the y-axis) and the transformation > H for kx > 0 0 kx (a dilation from the x-axis), the product

>

ky 0 1 0 ky 0 ky 0 1 0 H> H=> H> H=> H 0 kx 0 1 0 kx 0 1 0 kx

is a composite dilation, and this composition is commutative (it doesn’t matter in which order the transformations are applied, the final image is the same). x After such a composite dilation, y = f(x) becomes y = k x f e o. ky Reflections in lines through the origin -1 0 1 0 0 1 The matrices > H, > H and = G represent (i) a reflection in the y-axis, 1 0 0 1 0 -1 (ii) a reflection in the x-axis and a (iii) reflection in the line y = x respectively, since for any (x, y) we have: -x -1 0 x i > H= G = > H y 0 1 y x 1 0 x ii > H= G = > H 0 -1 y -y y 0 1 x G= G = = G 1 0 y x That these matrices are indeed the correct representations for the corresponding transformation can readily be seen by considering a general a c transformation matrix = G and choosing a, b, c and d so that the requisite b d transformation of coordinates applies. For example, reflection in the y-axis -x a b x maps the point (x, y) to the point (–x, y), or in matrix form = G = G = > H. c d y y Thus, as we require ax + by = –x, this can be obtained by having a = –1 and b = 0. Similarly, as we require cx + dy = y, this is obtained by having c = 0 -1 0 a b and d = 1, so the required transformation matrix is = H G=> c d 0 1 It is a useful exercise for students to similarly produce the other reflection matrices mentioned above, and apply the same reasoning to the dilation matrices covered earlier. iii =

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MathsWorks for Teachers Matrices Example 4.10

-1 0 What is the effect of the matrix > H corresponding to a reflection in 0 1 the y-axis on the graphs of the functions f and g? Solution

x1 -x -1 0 x Under this transformation > H = G = > H = > H, and so x = –x1 and y1 y 0 1 y y = y1. Thus, in general, y = f(x) is transformed to y1 = f(–x1) or y = f(–x). So y = x 2 is transformed to y = (–x)2 = x 2 and y = sin(x) is transformed to y = sin(–x) = –sin(x). When the image of a function or relation is the same as the original function under a transformation, the transformation illustrates a symmetry of the function or relation (see Leigh-Lancaster, 2006). In this case, for y = x 2, the symmetry exhibited is reflection (mirror) symmetry in the vertical coordinate axis. Example 4.11

1 0 H corresponding to a reflection in 0 -1 the x-axis on the graphs of the functions f and g?

What is the effect of the matrix > Solution

Under this transformation >

x1 x 1 0 x H = G = > H = > H, and so x = x1 and y1 0 -1 y -y

y = –y1. Thus, in general, y = f(x) is transformed to –y1 = f(x1) or y = –f(x). So y = x 2 is transformed to y = –(x)2 and y = sin(x) is transformed to y = –sin(x). Example 4.12

0 1 G corresponding to a reflection in 1 0 the line y = x on the graphs of the functions f and g? What is the effect of the matrix = Solution

Under this transformation =

x1 y 0 1 x G = G = = G = > H, and so x = y1 and y1 1 0 y x

y = x1. Thus, in general, y = f(x) is transformed to x1 = f(y1) or x = f(y). So y = x 2 is transformed to x = y2 and y = sin(x) is transformed to x = sin(y).

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The transformed functions are no longer functions but relations. In fact each function and its transformed relation are, by definition, inverse relations. For any one-to-one function h, the inverse relation is also a one-to-one function. Such functions are useful in solving equations by hand and using technology, since, for such a function, h, the equation h(x) = k will have the corresponding solution x = h–1(k). Reflection in the line y = mx To find the transformation matrix, we need to find the images of the point A with coordinates (1, 0) and the point B with coordinates (0, 1) under the transformation. We will assume in the following that 0 < m < 1. The reader can adapt the argument for other values of m. Consider the diagram shown in Figure 4.14. y P

1

y = mx Q θ –2

–1

O

θ

A 1

2

x

–1 Figure 4.14: Finding the image of A(1, 0) under reflection in the line y = mx, where 0 < m < 1

First we find the image of point A with coordinates (1, 0). The line through A perpendicular to the line y = mx cuts the line y = mx at Q and passes through the point P, where length of PQ = length of AQ. So P is the image of A after reflection in the line y = mx, and OP is also of length 1 unit. Let θ be the angle that the line makes with the positive x-axis, so tan (θ) = m. Then the angle POA = 2θ and so the coordinates of P are (cos(2θ), sin(2θ)) by definition. Next we find the image of point B with coordinates (0, 1), as shown in Figure 4.15.

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MathsWorks for Teachers Matrices y B 1 θ

–2

–1

O

y = mx

θ

R T 1

2

x

θ S –1 Figure 4.15: Finding the image of B (0, 1) under reflection in the line y = mx, where 0 < m < 1

The line through B perpendicular to y = mx intersects the line y = mx at R, cuts the x-axis at T, and S is the image of B after reflection in the line y = mx. Angles ORB and ORS are both right angles, and so, since the sum of angles in a triangle is 180°, the magnitude of angle TOS = 90° – 2θ. Hence the coordinates of S are (cos(–(90° – 2θ)), sin(–(90° – 2θ))). Using the double angle formulas

cos(A – B) = cos(A)cos(B) + sin(A)sin(B) and sin(A – B) = sin(A)cos(B) – cos(A)sin(B)

we see that the coordinates of S are (sin(2θ), –cos(2θ)). Hence a reflection in cos ]2qg sin ]2qg the line y = mx can be represented by the matrix > H, where sin ]2qg - cos ]2qg

p tan(θ) = m and -p 2 1 q 1 2 . In most cases it will not be possible to evaluate θ 2m 1 - m2 ] g cos 2 q = exactly, but sin ]2qg = and for a reflection in the 1 + m2 1 + m2 line y = mx.

Example 4.13

The graph of the function y = f(x) is reflected in the line y = 2x. Find the equation of the transformed function. Solution

Since m = tan(θ) = 2, it follows that θ = arctan(2), sin(2θ) = 0.8 and x1 - 0.6 0.8 x cos(2θ) = –0.6. Consider > H = G = > H. y 0.8 0.6 y 1

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As reflection transformations are self-inverse, the inverse of a reflection matrix will be the reflection matrix itself (which can be readily verified by using the standard form for the inverse of a - 0.6 0.8 x1 x 2 × 2 matrix), = G = > H > H. y 0.8 0.6 y1 Thus, the equation of the transformed function will be 0.8x + 0.6y = f(–0.6x + 0.8y). Rotations about the origin What is the matrix representing rotation about the origin in the anticlockwise direction through an angle θ, as shown in Figure 4.16? y 1

P

Q θ θ O

–1

x

1

Figure 4.16: Rotation of points (1, 0) and (0, 1) anticlockwise about the origin through an angle q 2 0

The point (1, 0) is rotated to point P with coordinates (cos(θ), sin(θ)) and the point (0, 1) is rotated to the point Q with coordinates (cos(90° + θ), sin(90° + θ)) = (–sin(θ), cos(θ)). Hence the matrix corresponding to an anti-clockwise rotation about the origin through the angle θ is:

=

cos ]qg - sin ]qg G sin ]qg cos ]qg

This result can be used to establish the compound angle formulas used in the previous section, by considering a rotation through an angle of θ1 + θ2 as both a single rotation through an angle of θ1 + θ2 and as a composition of two rotations, one through an angle of θ1 followed by another through an angle of θ2. By definition, the two corresponding matrices must be equal, so the corresponding elements give the required identities (see Leigh-Lancaster, 2006, pp. 66–8).

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MathsWorks for Teachers Matrices Example 4.14

Find the resultant function when the graph of y = f(x) is rotated anticlockwise about the origin through an angle of 60°. Find the image for the particular case when f(x) = 2x. Solution

The matrix corresponding to an anticlockwise rotation of 60° about the cos ]60cg - sin ]60cg origin is = G. sin ]60cg cos ]60cg Then =

x1 cos ]60cg - sin ]60cg x G= G = = G y1 sin ]60cg cos ]60cg y

-1 x1 so x cos ]60cg - sin ]60cg G > H = G== ] g ] g y sin 60c cos 60c y 1

cos ]60cg sin ]60cg x1 => H> H - sin ]60cg cos ]60cg y1 and y = f(x) is transformed to –xsin(60°) + y cos(60°) = f(xcos(60°) + y sin(60°)). Using the known exact surd values for sin(60°) and cos(60°) we x 3 y x y 3 obtain - 2 + 2 = f d 2 + 2 n . For the particular case f(x) = 2x, the x 3 y x y 3 image will be - 2 + 2 = 2 d 2 + 2 n , which can be rearranged to 5 3 8 give y =- x c 11 + 11 m . Students should be encouraged to explore these special cases: 1 When θ = 90° the rotation matrix becomes =

0 -1 G. 1 0

-1 0 2 When θ = 180° the rotation matrix becomes = G. 0 -1

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C o m p o s i t i o n o f l i n e ar t ra n s f o r m a t i o n s When several transformations are applied in a sequence, the matrix of the resulting transformation is simply the product of the corresponding matrices x applied from right to left. If = G is the coordinate vector of any point in the y x plane, and the linear transformation S is applied to = G then the resultant y x xl image coordinates will by given by = G = S = G. If a second transformation T is y yl applied to this, then the resultant image coordinates of the combined x x xl transformations will be given by T = G = T fS = Gp = ]TSg = G. The process is yl y y likewise repeated if further linear transformations are applied. Example 4.15

Find the matrix of the transformation consisting of a reflection in the y-axis followed by an anticlockwise rotation about the origin through an angle of 45°. Solution

-1 0 The matrix for the first transformation is > H, while the matrix for 0 1 R1 1 V S W ] g ] g cos 45c - sin 45c 2 2 W. the second transformation is = G= S 1 1 sin ]45cg cos ]45cg SS W 2 2 W T X To obtain the image of each point (x, y) in the plane under the given R1 1 V S 2 - 2 W -1 0 x W> sequence of these two transformations we use S 1 H = G. 1 SS WW 0 1 y 2 2 T X So the matrix acting upon each point is the product of the rotation matrix and the reflection matrix: R1 R 1 1 V 1 V S 2 - 2 W -1 0 S- 2 - 2 W W> W A=S 1 H= S 1 1 1 SS WW 0 1 SSW 2 2 2 2 W T X T X

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If the sequence of application of these two transformations is reversed, that is, we seek to find the matrix of the transformation consisting of a rotation about the origin through an angle of 45° followed by a reflection in the y-axis, then the combined transformation matrix will be R1 R 1 1 V 1 V S W S- 2 2 W -1 0 2 2 W S W, which is not the same as the previous A => = HS 1 1 1 0 1 SS WW SS 1 W 2 2 2 2W T X T X combined transformation matrix. The order of the transformations is important for the composition of these two transformations, unlike the case for composition of two rotations about the origin. In general, the matrices corresponding to the linear transformations should be multiplied from right to left matching the sequence of application of the transformations. Knowledge of the geometry of the transformations involved will provide insight into whether the composition of transformations in reverse order yields the same result as the composition of transformations in the original order, and consequently whether the matrix product is commutative. In general as matrix multiplication is not commutative, so the composition of transformations will not be the same when their sequence is reversed unless this is a consequence of the nature of the transformations involved.

S t u d e n t ac t i v i t y 4 . 4 a

Find the coordinates of the images of the vertices of the unit square under the



k 0 transformation with matrix > y H , and show that its area is given by kxky. 0 kx

b

Show that the rule of the image of y = f(x) under the transformation with matrix



k 0 x H , where k x ! 0 and k y ! 0 , is y = k x f e o . > y ky 0 kx

c

Find the image of the relation x2 + y2 = 1 (the unit circle) under the transformation



k 0 with matrix > y H where kx = b and ky = a and a ! 0 and b ! 0 . Hence find the 0 kx



formula for the area of the ellipse with horizontal axis length 2a and vertical axis length 2b.

d

Find the transformation matrix for an anticlockwise rotation about the origin through an angle θ followed by an anticlockwise rotation about the origin through an angle φ. Use the transformation matrix for an anticlockwise rotation about the origin through an angle θ + φ to show: i sin(θ + φ) = sin(θ)cos(φ) + cos(θ)sin(φ) ii cos(θ + φ) = cos(θ)cos(φ) – sin(θ)sin(φ)

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e f

The matrix =

1 k G represents a shear transformation in the x direction and the 0 1

1 0 matrix = G represents a shear transformation in the y direction. Find the image k 1 1 3 of the unit square under the shear transformation with matrix = G and draw the 0 1 original and its image on the same graph. Find the image of the function y = f(x) under the shear transformation with matrix =

1 3 G . In particular, find the image of y = x2 under this transformation. 0 1

A f f i n e t ra n s f o r m a t i o n s A function T: R2 → R2 is called an affine transformation if T(u) = Au + B, where u is an ordered pair (position vector) corresponding to a point in the plane, A is a 2 × 2 matrix and B is a fixed element of R2. A is called the matrix for T, and B the constant vector. It follows that every linear transformation is 0 an affine transformation, where B = ^0, 0h = = G. However, affine 0 transformations also include translations parallel to the coordinate axes, and combinations of these translations. With some discussion similar to the earlier case for linear transformations, it will be apparent to students that affine transformations also transform straight lines to straight lines and line segments to line segments. However, unlike linear transformations, they do 0 not necessarily transform the origin ^0, 0h = = G to itself, or straight lines 0 through the origin to straight lines through the origin. Teachers may wish to explore some simple examples with students to establish this observation, for example, finding the image of y = x under the affine transformation x 1 0 x 0 T f= Gp = = G = G + = G, and the natural generalisation. y 0 2 y 1 Translations parallel to the axes To translate a point p units in the positive direction of the x-axis, we would x x p use T f= Gp = I = G + = G, where I is the identity matrix of order 2. To translate a y y 0 point q units in the positive direction of the y-axis, we would use

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x x 0 T f= Gp = I = G + = G. We can form the composite of these transformations to q y y translate a point p units in the positive direction of the x-axis and q units in x x p the positive direction of the y-axis, by T f= Gp = I = G + = G. In this last case, y y q x1 x1 x1 - p x x p x p T f= Gp = I = G + = G = > H, so = G = > H - = G = > H and y = f(x) will be y1 y1 y1 - q y y q y q transformed to y - q = f^x - ph. Example 4.16

Find the image of the point (2, 6), the line y = 3x and the parabola y = x 2 under a translation by three units in the x-direction and two units in the y-direction. Solution

x 1 0 x 3 T f= Gp = = G= G + = G y 0 1 y 2 2 1 0 2 3 2 3 5 so T f= Gp = = G= G + = G = = G + = G = = G 6 0 1 6 2 6 2 8 Suppose (x1, y1) is the image of the point with coordinates (x, y). x1 x 1 0 x 3 Then T f= Gp = = G= G + = G = > H y1 y 0 1 y 2 x1 x1 - 3 x 3 Hence = G = > H - = G = > H y1 y1 - 2 y 2 Then x = x1 − 3 and y = y1 − 2, and the line y = 3x becomes the line y1 − 2 = 3(x1 − 3) or y = 3x − 7. For the case of the parabola y = x 2, we have y1 − 2 = (x1 − 3)2 or y = (x − 3)2 + 2.

C o m p o s i t i o n o f a f f i n e t ra n s f o r m a t i o n s If S and T are affine transformations, then so is the composite transformation T °S defined by T °S(u) = T(S(u)). Here S is applied to u first, and T is then applied to the result. Since u is a 2 × 1 coordinate (vector) matrix, S(u) is also a 2 × 1 coordinate (vector) matrix by conformability of matrix multiplication

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and addition. The same argument then applies for the application of T to the 2 × 1 coordinate matrix S(u). Example 4.17

a Find the image of the point u = (x, y) after a reflection in the y-axis followed by a dilation from the x-axis by a factor of 2. b Find the image of the point u = (x, y) after a dilation from the x-axis by a factor of 2 followed by a reflection in the y-axis. Solution

a Let S be a reflection in the y-axis, so S has the matrix representation -1 0 > H; and let T be a dilation from the x-axis by a factor of 2, so T 0 1 1 0 has the matrix representation = G. 0 2 -x 1 0 -1 0 x H = G = > H, where T °S has the G> 0 2 0 1 y 2y -1 0 1 0 -1 0 matrix representation = H=> H. G> 0 2 0 1 0 2 b In the reverse order of application we have -x -1 0 1 0 x S°T(u) = S(T(u)) = > H= G = G = > H, where S°T also has the 2y 0 1 0 2 y -1 0 1 0 -1 0 matrix representation > H= H. G=> 0 1 0 2 0 2 Thus composition of these two transformations is commutative, that is T °S = S°T.

Then T °S(u) =T(S(u)) = =

Students should be able to verify this for themselves by considering the geometric interpretation of both compositions. Example 4.18

The following sequence of affine transformations is applied to the region bounded by the unit circle {(x, y): x 2 + y2 = 1, –1 ≤ x ≤ 1} to obtain an obliquely oriented ellipse: 1 dilation by a factor of 3 from the y-axis and a factor of 2 from the x-axis

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2 rotation through an angle of 45° anti-clockwise about the origin 3 translation 1 unit parallel to the x-axis and 1 unit parallel to the y-axis Find the equation of the resulting ellipse, find its area and then draw the corresponding graph. Solution

x The dilations are applied by multiplying an arbitrary position vector = G y 3 0 by the matrix = G. The rotation is then applied by multiplying the 0 2 R V S 1 - 1 W 2 2W result by the matrix SS , and the translation is applied by 1 1 W S 2 W 2 T X 1 addition of the matrix = G. 1 R V S 1 - 1 W x1 1 2 2W3 0 x Hence, in combination, > H = SS G= G + = G = W 1 1 y1 0 2 y 1 S 2 2 W T X Reversing the order (that is, applying the inverse transformations) we obtain: R V S 1 - 1 W x1 1 2 2W3 0 x G= G > H - = G = SS = 1 1 W0 2 y y1 1 S 2 W 2 T RX V S 1 - 1 W x1 - 1 2 2W3 0 x which gives > H = SS G= G = 1 1 W0 2 y y1 - 1 S 2 2 W R V R T V X 1 1 W- 1 S x1 2 + y1 2 - 2 W -1 S x 3 0 S 2 2 W x1 - 1 6 3 W S 6 and so = G = = H= S G S > W. 1 1 W y1 - 1 y1 2 x1 2 y 0 2 S WW S 2 W S 2 4 4 T X T X Substituting these values for x and y into the equation for the unit circle x 2 + y2 = 1 yields the relation:

13x 2 − 2x(5y + 8) + 13y2 − 16y = 56 The area of the original region is π square units.

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The area of the transformed region is R V S 1 - 1 W 3 0 2 2W p # det = = 6p square units. G # det SS 1 1 W 0 2 S 2 W 2 T X The graph of both the original and the transformed relations are shown in Figure 4.17. y 4 3 2 1 –6

–5

–4

–3

–2

–1

O

1

2

3

4

5

6

x

–1 –2 –3 Figure 4.17: Composition of affine transformation from unit circle to an oblique translated ellipse

The processes of finding the rule of an (affine) transformed function or relation can be carried out readily using CAS, and the corresponding functions or relations graphed.

S t u d e n t ac t i v i t y 4 . 5 a

b

1 The graph of the function with rule f]x g = x is transformed as follows: 1 a dilation by a factor of 0.5 from the y-axis 2 a reflection in the y-axis 3 a translation of 3 units parallel to the x-axis and a translation of 1 unit parallel to the y-axis. Use matrices to find the rule of the transformed function. The graph of the relation x – y2 = 0 is reflected in the x-axis and then translated 2 units to the right and 1 unit down. Find the rule of the relation for the transformed graph.

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SUMM A R Y

• Matrices provide a natural representation for coordinate (position) x 1 0 vectors in the cartesian plane where ^x, yh = = G = x = G + y = G. y 0 1 1 0 The matrices (vectors) = G and = G are said to form a basis for the 0 1 coordinate vectors of the cartesian plane, since any coordinate vector can be written as a linear combination of these two vectors. • A linear transformation of the cartesian plane is a function T: R2 → R2, where T(x, y) = (ax + by, cx + dy) and a, b, c and d are real numbers. In matrix form this can be represented by: ax + by a b x H G= G = > = cx + dy c d y • Linear transformations (with non-singular matrices) map straight lines onto straight lines and preserve the parallel relation between straight lines. • The image of the origin under a linear transformation is itself, and a straight line passing through the origin is mapped onto another straight line passing through the origin (provided the transformation matrix is non-singular). a 1 0 • If the images of = G and = G under a linear transformation T are = G 0 1 c b x a b x and = G respectively, then T f= Gp = = G = G. y d c d y p r • If the images of = G and = G under a linear transformation T with q s a b p r a b matrix A are = G and = G respectively, then A = G = = G and q s c d c d a b p r -1 G # = G if the inverse matrix exists. q s c d • If P, with position vector p = (x1, y1), and Q, with position vector q = (x 2, y2) are two distinct points, and d = q – p = (x 2 – x1, y2 – y1), then the position vector r of any point on the line that contains P and Q is given by r = p + td, t ∈ R. A ==

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SUMM A R Y (Cont.)

• The linear function with rule y = mx + c can be written in matrix x 0 1 (vector) parametric form as = G = = G + t = G, where t ∈ R. c m y • To find the image of the graph of y = f(x) or f(x, y) = c under the x1 x linear transformation with matrix A, we substitute = G = A- 1 > H y1 y into y = f(x) or f(x, y) = c, if A–1 exists. ky 0 •> H where ky and kx are positive real numbers, represents a 0 kx dilation by a factor ky from the y-axis and a dilation by a factor kx cos ]2qg sin ]2qg from the x-axis in either order; > H represents sin ]2qg - cos ]2qg

cos ]qg - sin ]qg G sin ]qg cos ]qg represents a rotation through an angle θ anticlockwise about the a origin; and = G represents translation by vector (a, b) (a units in the b x-direction and b units in the y-direction when a 2 0 and b 2 0). • An affine transformation of the cartesian plane is a function T: R2 → R2 such that T(u) = Au + B. In matrix form this can be ax + by + e e a b x represented by = H. Affine G= G + = G = > cx + dy + f f c d y reflection in the line y = mx where m = tan(θ); =

transformations (where A–1 exists) map straight lines onto straight lines and preserve the parallel relation between straight lines; however, lines through the origin are not necessarily mapped to other lines through the origin. • If S and T are affine transformations, then so is the composite transformation T °S defined by T °S(u) = T(S(u)). The composition may, or may not, be commutative. This may be determined by computation of the respective matrices, or by geometric interpretation of the transformations involved.

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SUMM A R Y (Cont.)

• When a translation is part of an affine transformation, this may occur before or after composition with other linear transformations. – If the translation component occurs last, then the affine transformation has the form X1 = AX + B, so X = A–1(X1 – B) will determine x and y in terms of x1 and y1 for substitution into the rule of the original function or relation to determine the rule of the transformed relation or function. – If the translation component occurs first, then the affine transformation has the form X1 = A(X + B), so X = A–1 X1 – B will determine x and y in terms of x1 and y1 for substitution into the rule of the original function or relation to determine the rule of the transformed function or relation. References Anton, H & Rorres, C 2005, Elementary linear algebra (applications version), 9th edn, John Wiley and Sons, New York. Cirrito, F (ed.) 1999, Mathematics higher level (core), 2nd edn, IBID Press, Victoria. Evans, L 2006, Complex numbers and vectors, ACER Press, Camberwell. Leigh-Lancaster, D 2006, Functional equations, ACER Press, Camberwell. Nicholson, KW 2003, Linear algebra with applications, 4th edn, McGraw-Hill Ryerson, Whitby, ON. Sadler, AJ & Thorning, DWS 1996, Understanding pure mathematics, Oxford University Press, Oxford.

Websites http://wims.unice.fr/wims/en_tool~linear~matrix.html – WIMS This website provides a matrix calculator. http://en.wikipedia.org/wiki/Linear_transformation – Wikipedia This site provides a comprehensive discussion on linear transformations with links to other resources and references. http://en.wikipedia.org/wiki/Affine_transformation – Wikipedia This site provides a comprehensive discussion on affine transformations with links to other resources and references. http://www.ies.co.jp/math/java/misc/don_trans/don_trans.html This website contains an applet that shows how the shape of a dog is transformed by a 2 × 2 matrix. http://merganser.math.gvsu.edu/david/linear/linear.html This website contains an applet that allows you to move a slider adjusting coefficients in a 2 × 2 matrix and see the effect of the equivalent transformation on the unit square.

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5

Tra n s i t i o n m a t ric e s Conditional probability One of the key ideas that students come across early in their study of probabilities related to compound events for a given event space is the notion of conditional probability and the associated ideas of dependent and independent events. While students’ experience with subjective probability makes them well acquainted with events that may or may not be dependent, such as the likelihood of scoring on the second of two free shots in a basketball game, given that one may or may not have scored on the first shot, their school study of probability often begins with compound events that are (physically) independent events. On the other hand, a combination of experience and knowledge indicates that certain events are dependent, for example, gender and colour blindness to red and green, or having a disease and the likelihood of testing positive for the disease. Whether events in a given context are independent or not, is not always clear. Thus, experiments with tossing coins and rolling dice, which are physically independent, can lead to an implicit willingness, or preference, to assume that two events from a given event space are independent. One of the more intriguing and counterintuitive scenarios involving conditional probability is the game show problem called the Monty Hall dilemma (see the UCSD website). This is a good context for stimulating student interest, many people think the events involved are (or should be) independent. If we consider two events A and B from the same event space, U, then the conditional probability of A given B, that is the probability that event A occurs given that event B has occurred, written as Pr ^ A | Bh, corresponds to the proportion of B events that are also A events, relative to the proportion of B Pr ]A k Bg events; that is, Pr ^ A | Bh = . Conditional probability can be used to Pr ]Bg discuss both dependence and independence of events.

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If A and B are independent events, then Pr ^ A | Bh will be the same as Pr(A) Pr ]A + Bg and so in this case Pr ^ A | Bh = = Pr ]Ag or Pr ]Bg Pr ]A + Bg = Pr ]Ag # Pr ]Bg. Similarly, Pr ^B | Ah will be the same as Pr(B) and Pr ]A + Bg so in this case Pr ^B | Ah = = Pr ]Bg or, as before, Pr ]Ag Pr ]A + Bg = Pr ]Ag # Pr ]Bg. If Pr(A | B) is different from Pr(A), and likewise Pr(B | A) will be different from Pr(B), then A and B are dependent events. Any two events A and B from a given event space may or may not be dependent; however, the following relationships (sometimes called the law of total probability for two events) holds irrespective of whether this is the case or not: Pr ]Ag = Pr ^ A | Bh # Pr ]Bg + Pr ^ A | Blh # Pr ]Blg Pr ]Bg = Pr ^B | Ah # Pr ]Ag + Pr ^B | Alh # Pr ]Alg

where A′ and B′ are the complements of A and B in U respectively. In practice, these relationships can be represented using a Venn diagram or Karnaugh map, a tree diagram, or by matrices. It is important to ensure that students are familiar with each of these representations and their use, to assist in solving problems related to probabilities associated with simple compound events.

Tra n s i t i o n p r o b a b i l i t i e s Many natural systems undergo a process of change where at any time the system can be in one of a finite number of distinct states. For example, the weather in a city could be sunny, cloudy and fine, or rainy. Such a system changes with time from one state to another, and at scheduled times, or stages, the state of the system is observed. At any given stage, the state to which it changes at the next stage cannot be determined with certainty, but the probability that a given state will occur next can be specified by knowing the current state of the system. That is, the probability that the system will be in a given state next is conditional only on the current state. Such a process of change is called a Markov chain or Markov process. The conditional probabilities involved (that is, the probabilities of going to one state given that the system was in a certain state) are called transition probabilities, and the process can be modelled using matrices.

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We begin by considering the following example. Jane always does the weekly shopping at one of two stores, A and B. She never shops at A twice in a row. However, if she shops at B, she is three times as likely to shop at B the next time as at A. Suppose that she initially shops at A. This is an example of a Markov chain, since the store at which she shops next depends only on the store she shopped at the week before, and the conditional probabilities for each possible outcome are the same on each occasion. There are two states, state 1 which corresponds to shopping at store A, and state 2 which corresponds to shopping at store B. We can represent this by a tree diagram, as shown in Figure 5.1 and set up a corresponding table of transition probabilities, as shown in Table 5.1. Current 0

Next A

A 1

B A

0.25 B 0.75

B

Figure 5.1: Tree diagram representation for the first transition

Table 5.1: Summary of transition probabilities

Present week’s store

Next week’s store

A

B

A

0

0.25

B

1

0.75

Note that the columns of the table sum to 1. The store at which Jane shops in a given week is not determined. The most we can expect to know is the probability that she will shop at A or B in that week. Let s(1m) denote the probability that she shops at A in week m, and s(2m) the probability that she shops at B in week m.

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Use the law of total probability to find the probability that Jane shops at store A in week 1 and the probability that Jane shops at store B in week 1 given that she shops at store A initially. Solution

As she shops at A initially, s(10) = 1 and s(20) = 0. For the next week (using the law of total probability): s(11) = Pr ^shops at A in week 1 | shopped at A in week 0h

# Pr ^shopped at A in week 0h + Pr ^shops at A in week 1 | shopped at B in week 0h # Pr (shopped at B in week 0) = 0 # 1 + 0.25 # 0 = 0

s(21) = Pr ^shops at B in week 1 | shopped at A in week 0h

# Pr ^shopped at A in week 0h + Pr ^shops at B in week 1 | shopped at B in week 0h # Pr (shopped at B in week 0) = 1 # 1 + 0.75 # 0 = 1

If we write S0 = >

s]10g

s]11g s]1mg S = S = and , then, in general, > ]1gH > ]mgH is H 1 m s]20g s2 s2

called the state vector for week m, since it gives the probabilities of being in any state after m weeks (or transitions). It is convenient to let S0 correspond to the initial week. Since the definition of matrix multiplication corresponds naturally to the calculations we wish to carry out for this purpose, these calculations can be written in matrix form as follows: S1 = >



s]11g

H==

s]21g

0 0.25 1 0 G = G = PS0 = = G 1 0.75 0 1

0 0.25 G is the matrix of transition probabilities, and is called 1 0.75 the transition matrix. Teachers should take care to ensure that students follow the modelling process and make the conceptual connections between the law of total probability, its application to the transition state problem, and the subsequent representation using matrices and their products. where P = =

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Hence, from S1 we can see that given that she shopped at A in week 0, the probability that she shops at A in week 1 is 0 and the probability that she shops at B in week 1 is 1. What happens two weeks after shopping at A initially? If we use a tree diagram representation, as shown in Figure 5.2, we can calculate the corresponding probabilities for the first two transitions. Initially 0 A

Week 1 A

1 B

0 1

Week 2 A

0.25

B A

0.75

B

Figure 5.2: Tree diagram representation for the first two transitions

s(12) = Pr ^shops at A in week 2 | shopped at A in week 1h

# Pr ^shopped at A in week 1h + Pr ^shops at A in week 2 | shopped at B in week 1h # Pr (shopped at B in week 1) = 0 # 0 + 0.25 # 1 = 0.25

s(22) = Pr ^shops at B in week 2 | shopped at A in week 1h

# Pr ^shopped at A in week 1h + Pr ^shops at B in week 2 | shopped at B in week 1h # Pr (shopped at B in week 1) = 1 # 0 + 0.75 # 1 = 0.75

In matrix terms, this is equivalent to S2 = >

s]12g

H==

s]22g

0 0.25 0 0.25 G = G = PS1 = = G. 1 0.75 1 0.75

Hence, given that Jane shopped at A in week 0, the probability that she shops at A in week 2 is 0.25 and the probability that she shops at B in week 2 is 0.75. 0.25 0.1875 1 0.25 Moreover, S2 = PS1 = P^PS 0h = P2 S0 = = G= G = = G (the first 0.75 0.8125 0 0.75 column of P2). This can be extended to week 3; however, the process of calculation using tree diagrams becomes increasingly time and space consuming, whereas the

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matrix form offers a much more convenient representation for these calculations, where S 3 = PS2 = P 3 S0, and in general, for m weeks later: Sm = PSm - 1 = Pm S0



Where many transitions may take place the relevant matrix calculations are best done by technology, using, for example, a CAS. Example 5.2

Find the probabilities that Jane shops at A (i) 3, (ii) 4, (iii) 5, (iv) 10, (v) 50 and (vi) 100 weeks later. Solution

i ii iii iv v vi

0 0.25 0.25 0.1875 G= G== G, and so the probability that she 1 0.75 0.75 0.8125 shops at A three weeks later is 0.1875. 0 0.25 0.1875 0.203125 S 4 = PS 3 = = G= G== G, and so the probability 1 0.75 0.8125 0.796875 that she shops at A 4 weeks later is 0.203125 0.199 S5 = P5 S0 . = G, and so the probability that she shops at A 0.801 5 weeks later is approximately 0.199. 0.200 S10 = P10 S0 . = G, and so the probability that she shops at A 0.800 10 weeks later is approximately 0.200. 0.200 S 50 = P 50 S0 . = G, and so the probability that she shops at A 0.800 50 weeks later is approximately 0.200. 0.200 S100 = P100 S0 . = G, and so the probability that she shops at A 0.800 100 weeks later is approximately 0.200. We can note that 0.2 0.2 P100 . = G; in fact there seems to have been very little change, 0.8 0.8 at this level of accuracy for larger values of Pm. S 3 = PS2 = =

In this example, the state vectors S0 , S1, S2 ... Sn appear to converge to 0.2 S = = G as n increases. 0.8

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If this is indeed the case, then we can say that the long-term probability of Jane shopping at A is 0.2 and shopping at B is 0.8. That is, in the long term she will shop at A 20% of the time and at B 80% of the time, provided that she doesn’t change her pattern of behaviour in this regards. If instead of shopping at A in week 0, we knew she shopped at B in 0 week 0, then S0 = = G, and the corresponding probabilities are shown in 1 Example 5.3. Example 5.3

Find the probabilities that Jane shops at A in (i) 1, (ii) 2, (iii) 3, (iv) 4, (v) 5, (vi) 10, (vii) 50 and (viii) 100 weeks later, given that she initially shopped at B. Solution

i ii iii iv v vi

0 0.25 0 0.25 G= G = = G, and so the probability that she 1 0.75 1 0.75 shops at A 1 week later is 0.25. This state vector is the same as that after two transitions if she shops at A initially. 0 0.25 0.25 0.1875 S2 = PS1 = = G= G== G, and so the probability that she 1 0.75 0.75 0.8125 shops at A 2 weeks later is 0.1875. 0 0.25 0.1875 0.203125 S 3 = PS2 = = G= G== G, and so the probability 1 0.75 0.8125 0.796875 that she shops at A 3 weeks later is 0.203125. 0.199 S 4 = P 4 S0 . = G, and so the probability that she shops at A 0.801 4 weeks later is approximately 0.199. 0.200 S5 = P5 S0 . = G, and so the probability that she shops at A 0.800 5 weeks later is approximately 0.200. 0.200 S10 = P10 S0 . = G, and so the probability that she shops at A 0.800 10 weeks later is approximately 0.200. S1 = PS0 = =

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0.200 G, and so the probability that she shops at A 0.800 50 weeks later is approximately 0.200. 0.200 viii S100 = P100 S0 . = G, and so the probability that she shops at A 0.800 100 weeks later is approximately 0.200. vii S 50 = P 50 S0 . =

The long-term probabilities are as in the previous case, in which we assumed Jane initially shopped at A. So it appears that, with this pattern of shopping behaviour, in the long term Jane will shop at A in any week with probability 0.2, and at B in any week with probability 0.8, regardless of where she shopped initially. Table 5.2 summarises the state vectors for five transitions given that Jane either shopped at A or B initially, as well as the transition matrix raised to the number of transitions, for the first five transitions. Table 5.2: Summary of state and transition matrices for five transitions from either initial state

Pn , where n is the number of transitions

Number State vector given State vector given of she shops at A she shops at B transitions initially initially

0 = G 1

1

=

2

=

3

=

0.25 G 0.75

=

0.25 G 0.75

=

0.1875 G 0.8125

0.1875 G 0.8125

=

0.203125 G 0.796875

= =

0 0.25 G 1 0.75

0.25 0.1875 G 0.75 0.8125

0.1875 0.203125 G 0.8125 0.796875

4

=

0.203125 G 0.796875

=

0.199219 G 0.800781

=

0.203125 0.199219 G 0.796875 0.800781

5

=

0.199219 G 0.800781

=

0.200195 G 0.799805

=

0.199219 0.200195 G 0.800781 0.799805

We can see from this table that the columns of Pn contain the state vectors (after n transitions) after initially shopping at A or at B respectively. So the (i, j)th element of Pn gives the probability of starting in state j and moving to state i after n transitions.

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chapter 5 Transition matrices S t u d e n t ac t i v i t y 5 . 1 a b

Suppose initially we are unsure of where Jane shops, but it could be at either A or 0. 5 B with equal probability. Then S0 = = G . Find S1, S2, S5, S10, and S50. 0. 5 Consider a Markov chain, with two states 1 and 2, with transition probability matrix P, with pij = probability of going from state j to state i in one transition. 0.445 0.444 G . What is the probability of 0.555 0.556 going from state 1 to state 2 in three transitions? going from state 2 to state 2 in three transitions?

Suppose P3 = = i ii

T h e s t e a dy- s tat e v e c t o r It appears that in the long term, after many transitions, the state vectors converge to the same vector regardless of where Jane initially shopped. Such a vector is called a steady-state vector. If this is the case, how can we find the steady-state vector for this problem? We can phrase this question more generally, and suppose that P is the transition matrix of a Markov chain, and assume that the state vectors Sm converge to a limiting state vector S. Then Sm is very close to S for sufficiently large m, so Sm + 1 is also very close to S. Then the equation

Sm + 1 = PSm

is closely approximated by

S = PS

where S is a solution to this matrix equation. It is easily solved as it can be written as a system of linear equations in matrix form

(I – P)S = O

where the entries of S are the unknowns and I is the identity matrix. This homogeneous system has many solutions; the one we are most interested in is the one whose entries sum to 1.

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MathsWorks for Teachers Matrices Example 5.4

Find the steady-state vector S for Jane’s shopping. Solution



I-P ==

1 - 0.25 1 0 0 0.25 H G-= G=> 0 1 1 0.75 - 1 0.25

1 - 0.25 Using Gaussian elimination, this reduces to > H, and so the 0 0 s1 solution for S = > H is s1 = 0.25 × s2 with s1 + s2 = 1, hence s1 = 0.2, s2 s2 = 0.8. It is a natural question to ask if we can always find a steady-state vector, and whether the powers of the transition matrix converge to a matrix whose columns are equal to the steady-state vector. To answer this we need to introduce the notion of a regular transition probability matrix. We say that a transition probability matrix P is regular if, for some positive integer m, the matrix Pm has no zero entries. It can be shown that, if P is a regular transition probability matrix, then it has a unique steady-state vector S (see, for example, the Iowa State maths website). Further, the matrix defined by L = lim Pm exists, and is given by L = 6S | S | f | S@, that is a matrix where m"3 each column is a copy of the steady-state vector S, and if L = [lij] then lij is the long-term probability of being in state i if the system began in state j. It is straightforward to determine the steady state vector for a 2 × 2 transition probability matrix, given that it exists. To do this we use a general formulation, taking a transition probability matrix P = =

1-a b G, where a 1-b 0 # a # 1 and 0 # b # 1, and solve (I − P)S = O for S where I is the 2 × 2 identity matrix. Then 1 0 1-a b a -b G-= = G== G a 1 b a b 0 1 which reduces by Gaussian elimination to a -b = G. 0 0 x Writing S = = G, we have ax = by. There are now two possible cases: y ax ax = 1, and so Case 1: If b 2 0, then y = , and as x + y = 1, x + b b ]a + bg a a b = 1. Since b 2 0, x = and hence y = . x b1 + l = 1 or x b a+b b a+b

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Note that if a = 0, then y = 0 and x = 1, and state 1 is called an absorbing state. This means that once the system is in state 1 it will never leave it. Note 0 1 0 1 0.5 0.5 also that if a = 1 and b = 1, then P = = G is not regular but = G = G = = G, 1 0 1 0 0.5 0.5 and so the state vectors will converge if and only if the system has initial state 0.5 vector = G; that is, there is no steady state vector. 0.5 Case 2: If b = 0, then ax = 0, so a = 0 or x = 0. If a = 0, then the transition matrix is the identity matrix, so the system stays in whatever state it is in initially. So there is no steady state vector. If a ≠ 0 and x = 0, then y = 1, and state 2 is also an absorbing state. 1-a b In summary, if the transition probability matrix P = = G is a 1-b R V S b W a + bW regular, then the steady state vector is given by S = SS . a W Sa + b W T X S t u d e n t ac t i v i t y 5 . 2 a

1

Show that the transition probability matrix P = > 21 2

b c

1 H is regular. Find the 0

steady-state vector S and the limit matrix L = lim P n . n"3

0 1 G is not regular. Does the limit 1 0 matrix L = lim P m exist? Does it have a steady-state vector? m"3 a 0 For 0 1 a 1 1, show that the matrix P = = G is not regular. Find a 1-a 1 Show that the transition probability matrix P = =

steady-state vector and lim P n if it exists. n"3

A p p l ica t i o n s o f t ra n s i t i o n m a t ric e s Examples such as the following, and others from various practical and research contexts, or from the literature, can be used to help students develop the formulation, solutions and interpretation skills associated with modelling and problem solving that employs transition matrices and Markov chains. Keys aspects of these processes are: • consideration of features of the context that indicate a Markov process is likely to provide a suitable model

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• identification of relevant states and transition (conditional) probabilities and initial state (or states) • formulation of the transition matrix and initial state vector, and computation of relevant powers of the transition matrix and subsequent state vectors • analysis of long-run behaviour, including investigation and interpretation of possible steady-state or other behaviour of the system While the first few transitions for a two-state system, and its long-run behaviour, assuming convergence to a steady state, can readily be computed by hand calculation, student familiarity with the use of a suitable technology such as CAS is required for computation of higher powers in two-state problems, and problems where there are more than two states. Example 5.5

OzBank offers customers two choices of credit card: Ordinary and Gold. Currently 70% of its customers have an Ordinary card and 30% have a Gold card. The bank wants to increase the percentage of its customers with a Gold card, as it gets higher fees from these customers, and so sends out an offer to all Ordinary cardholders offering a free upgrade to a Gold card for twelve months. It expects that each month for the next three months, 10% of its Ordinary cardholders will upgrade to a Gold card, but 1% of Gold cardholders will downgrade to an Ordinary card. What percentage of its customers would have Gold cards at the end of the three months? Solution

This information can be summarised as a table: Current

One month later

Ordinary

Gold

Ordinary

0.90

0.01

Gold

0.10

0.99

The corresponding transition matrix is =

vector =

122

0.70 G. 0.30

0.90 0.01 G, with initial state 0.10 0.99

chapter 5 Transition matrices

To find the percentages three months later, we calculate 0.90 0.01 3 0.70 0.52 G = G.= G. 0.10 0.99 0.30 0.48 So after three months the bank could expect approximately 48% of its customers to have a Gold card. We can also observe that if the number of customers was fixed 700 during this period at, say, 1000, then we could use = G instead of the 300 700 0.70 initial state vector, since = G = 1000 # = G, and use this to calculate 300 0.30 the numbers in each state after each transition. This only works if the total number of objects remains constant over transitions.

=

Example 5.6

A wombat has its burrow beside a creek and each night it searches for food on either the east or west side of the creek. The side on which it searches for food each night depends only on the side on which it searched the night before. If the wombat searches for food on the east side one night, then the probability of the wombat searching on the east side of the creek the next night is 0.2. The transition matrix for the probabilities of the wombat searching for food on either side of the creek given the side searched on the previous night is

=

0.2 0.7 G 0.8 0.3

a If the wombat searches for food on the west side one night, what is the probability that it searches for food on the west side the next night? b If the wombat searches for food on the west side on the Monday night, what is the probability it searches for food on the west side again on the following Saturday (5 days later)? c In the long term, what proportion of nights will it spend searching for food on the west side?

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MathsWorks for Teachers Matrices Solution

a We can view the transition matrix as below. It is clear that if it searches on the west side one night then the probability that it searches for food on the west side the next night is 0.3. Side searched for food current night

Side searched for food the next night

East

West

East

0.2

0.7

West

0.8

0.3

R V S 77 W 0.2 0.7 0 160 W 0.48125 == b We need to find = G = G= S G, or alternatively, S 83 W 0.51875 0.8 0.3 1 S 160 W T X 5 0.2 0.7 0.45 0.48125 simply calculate = G == G. The probability the 0.8 0.3 0.55 0.51875 wombat searches for food on the west side on the following Saturday 83 is 160 = 0.51875. c Using the formula for the steady-state solution, with a = 0.8 and 0.7 7 b = 0.7, x = 0.7 + 0.8 = 15 and 0.8 8 y = 0.7 + 0.8 = 15. Hence in the long term the wombat will spend 8 15 of the nights searching for food on the west side. Alternatively, we can solve the matrix equation 1 0 0.2 0.7 x 0 f= G-= Gp = G = = G for x and y, with x + y = 1. 0 1 0.8 0.3 y 0 5

0.8 - 0.7 H using Gaussian 0 0 elimination (replacing row 2 by row 2 + row 1), and so 0.7y 7y 7y 0.8x – 0.7y = 0. Then x = 0.8 = 8 . As x + y = 1, 8 + y = 1, 15y 8 7 8 = 1 and so y = 15 and x = 15. Hence in the long term the 8 wombat will spend 15 of the nights searching for food on the west side. The coefficient matrix reduces to >

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chapter 5 Transition matrices

Also, we could take suitable large powers of P and observe whether there is significant change in the resultant values or not. 0.466667 0.466667 0.466667 0.466667 Now P20 . = G, and P 30 . = G, 0.533333 0.533333 0.533333 0.533333 so we appear to have convergence to 6 decimal places, and so the wombat will spend approximately 0.466667 (or just under 47%) of the nights searching for food on the east side and 0.533333 (or just over 53%) of the nights searching for food on the west side. Example 5.7

A wombat has its burrow beside a creek and each night it searches for food on either the other side of the creek or north or south of its burrow on the same side of the creek. The area in which it searches for food each night depends only on the area in which it searched for food the night before. If the wombat searches for food on the other side of the creek one night, then the probabilities of the wombat searching on the other side of the creek, or north or south of its burrow the next night are 0.2, 0.4 and 0.4 respectively. If the wombat searches for food north of its burrow one night, then the probability that it will search for food north of its burrow the next night is 0.1. The transition matrix for the probabilities of the wombat searching for food in each area given the area searched for food on the previous night is R V S0.2 0.5 0.5 W S0.4 0.1 0.3 W SS W 0.4 0.4 0.2 W T X a If the wombat searches for food on the south side of its burrow one night, what is the probability that he searches for food on the north side the next night? b If the wombat searches for food on the north side of its burrow on the Monday night, what is the probability it searches for food on the north side of its burrow again on the following Saturday (5 days later)? c In the long term, what proportion of nights will the wombat spend searching for food north of its burrow?

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MathsWorks for Teachers Matrices Solution

a We can view the transition matrix as below, and it is clear that if it searches for food on the south side one night then the probability that he searches for food on the north side the next night is 0.3. Side searched for food current night

Side searched for food the next night

Other

North

South

Other

0.2

0.5

0.5

North

0.4

0.1

0.3

South

0.4

0.4

0.2

R V 7711 R V5 R V S 20000 W R V S0.2 0.5 0.5 W S 0 W S 28101 W S 0.38555 W b We need to find S0.4 0.1 0.3 W S 1 W = SS 100000 WW . S 0.28101 W, and so SS W S W S W 0.4 0.4 0.2 W S 0 W S 1042 W S0.33344 W T X T X SS 3125 WW T X T X the probability it searches for food north of its burrow on the 28101 following Saturday is 100000 = 0.28101. JR1 0 0 V R0.2 0.5 0.5 VN Rx V R0 V W S WOS W S W KS c Solve K S0 1 0 W- S0.4 0.1 0.3 WOSy W = S 0 W for x, y and z, with K SS0 0 1 WW SS0.4 0.4 0.2 WWOSS z WW SS 0 WW LT X T XP T X T X x + y + z = 1. V R V R V R S1 0 0 W S0.2 0.5 0.5 W S 0.8 - 0.5 - 0.5W S0 1 0 W- S0.4 0.1 0.3 W = S- 0.4 0.9 - 0.3 W W SS W S W S 0 0 1 W S0.4 0.4 0.2 W S- 0.4 - 0.4 0.8 W T X T X TV X R S1 0 - 15 W 13 W S - 11 W S which reduces to S0 1 13 W by Gaussian elimination. S0 0 0 W T X 15z 11z Hence we have x = 13 and y = 13 with x + y + z = 1. This 1 11 5 gives 3z = 1, so z = 3 , y = 39 and x = 13 . Thus in the long term 11 the wombat will spend 39 (≈ 0.28205) of nights searching for food north of its burrow.

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Alternatively, consider suitably large powers of the transition matrix: R V S0.38462 0.38462 0.38462 W P15 . S0.28205 0.28205 0.28205 W SS W 0.33333 0.33333 0.33333 W T R X V S0.38462 0.38462 0.38462 W and P20 . S0.28205 0.28205 0.28205 W SS W 0.33333 0.33333 0.33333 W T X R V S0.38462 0.38462 0.38462 W and so we have P3 . S0.28205 0.28205 0.28205 W SS W 0.33333 0.33333 0.33333 W T X Hence the wombat will spend 0.28205 approximately of nights searching for food north of its burrow. Example 5.8

Consider a simple genetic model, involving just two types of alleles, A and a, for a gene. Suppose that a physical trait such as eye colour is controlled by a pair of these genes, one inherited from each parent. An individual could then have one of three combinations of alleles of the form AA, Aa or aa. A person may be classified as being in one of three states: Dominant (type AA): gene of type A from both parents Hybrid (type Aa): gene of type A from one parent and gene of type a from the other parent Recessive (type aa): gene of type a from both parents Assume that the gene inherited from a parent is a random ‘choice’ from the parent’s two genes and that each each parent is equally likely to transmit either of its two genes to an offspring. We can form a Markov chain by starting with a population and always crossing with hybrids to produce offspring. The time required to produce a subsequent generation is the time period for the chain. What is the corresponding transition matrix? Suppose we start with a person with dominant trait—type AA—and cross with a person with hybrid trait—type Aa. Type AA will always contribute A to the offspring, and type Aa will contribute A one half of the time and a one half of the time. If we start with a hybrid, and cross with a hybrid, we have the following situation: the first hybrid will contribute either A or

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a to the offspring, each with probability one-half. The second hybrid will also contribute either A or a to the offspring, again each with probability one-half. Hence we have one-quarter probability of AA, onequarter probability of aa and one-half probability of hybrid Aa. The transition matrix is as follows: R D H RV 1 1 D S2 4 0W S W 1 1 1W S H S2 2 2 W= P S W R S0 1 1 W 4 2 T X a What proportion of the third generation offspring (that is, after two time periods) of the recessive population has the dominant trait? b What proportion of the third generation offspring (after two time periods) of the hybrid population is not hybrid? c If, initially, the entire population is hybrid, find the population vector in the next generation. d If, initially, the population is evenly divided among the three states, find the population vector in the third generation (after two time transitions). e Show that this Markov chain is regular, and find the steady-state population vector. Solution

a We need to calculate P2, and obtain the number in the (1, 3) position. R V S3 1 1W S8 4 8 W 1 1 1 2 P = SS 2 2 2 WW S1 1 3W SS 8 4 8 WW T X Hence one-eighth of the third generation offspring of the recessive population has the dominant trait. b From P2, we see that one-half of the third generation offspring of the hybrid population has the hybrid trait, while one-quarter has dominant and one-quarter has the recessive trait. So one-half is not hybrid.

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chapter 5 Transition matrices

R V S1W R V S4W S0 W 1 c Here S0 = S 1 W, and so PS0 = SS 2 WW and this is the population SS WW S1W 0 T X S4W T X distribution vector for the next period. R V R V S1W S1W S3W S4W 1 1 d Here S0 = SS 3 WW, and so P2 S0 = SS 2 WW is the population vector for the S1W S1W SS 3 WW S4W T X T X third generation. This seems familiar. e As P2 has no zero entries, P is regular. If we approximate a suitably R V S0.25 0.25 0.25W high power of P, say P20, we get S 0.5 0.5 0.5 W. So we can guess SS W R V 0.25 0.25 0.25W S0.25W T X that the steady-state population distribution should be S 0.5 W. This SS W should be checked. 0.25W T X Begin with I - P: R V R V 1 1 1 1 R V S2 4 0W S 2 - 4 0 W W S1 0 0 W S 1 1 1 W S 1 1 1W W = SS0 1 0 W - S 2W SS W S2 2 2W S 2 2 0 0 1W S 1 1 W S 1 1 W T X S0 4 2W S 0 - 4 2 W T X T X and use technology to obtain the reduced row echelon form: R V S1 0 - 1 W S0 1 - 2 W S W S0 0 0 W R V T X Sx W Let S = Sy W. From the above, x = z and y = 2z. Then since SS WW z R V 0.25W T X S 1 x + y + z = 1, z = 4 and so S = S 0.5 W. SS W 0.25W T X At this stage it is useful to recall that a state in a Markov chain is called an absorbing state if it is not possible to leave that state over the next time period. If state i is an absorbing state, then in the ith column of the transition

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matrix, there will be a 1 in the ith row and zeroes everywhere else. When we take powers of the transition matrix, the ith column will remain the same, and so the transition matrix is not regular, and may not have a steady-state vector. Now, suppose that we always cross with recessives. R V S0 0 0 W S 1 W Then the transition matrix is P = S1 2 0 W. This has an absorbing state, S 1 W SS0 W 2 1W T X P, and see what happens. the recessive state. Let us take some powers of R V S 0 0 0W S1 1 W P2 = S 2 4 0 W S1 3 W SS 4 1 WW 2 T X R V 0 0 0 S W S1 1 W 3 P = S 4 8 0W S3 7 W SS 4 8 1 WW X V R T 0 0W S 0 S 1 W 1 P10 = S 512 1024 0 W S 511 1023 W SS W 512 1024 1 W T X Continuing with higher powers, it would appear that there will be a R V S0 W steady-state vector, equal to S 0 W. In the long term, we will end up with SS WW recessives. 1 R V T X 0 S W Check that S = S 0 W is a steady-state vector; that is, show PS = S. SS WW 1 T X Example 5.9

Humans have two sets of chromosomes, one obtained from each parent, which determine their genetic makeup. In this example we investigate the inbreeding problem.

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chapter 5 Transition matrices

Assume that two individuals mate randomly. In the next generation, two of their offspring of opposite sex mate randomly. Suppose the process of brother and sister sibling mating continues each generation. We can regard this as a Markov chain whose states consist of six mating states: State 1: AA × AA State 2: AA × Aa State 3: AA × aa State 4: Aa × Aa State 5: Aa × aa State 6: aa × aa Let P = [pij] be the corresponding transition matrix. Suppose that the parents are both of type AA. Then all children will be of type AA, and so a mating of brother and sister will only give AA. Hence p11 = 1. Suppose that the parents are of type AA × Aa. Then half their children will be of type AA and half will be of type Aa. A mating of these offspring will give 0.25 of AA × AA, 0.5 of AA × Aa and 0.25 of Aa × Aa, and so p21 = 0.25, p22 = 0.5 and p24 = 0.25. Continuing in this way, we can obtain Table 5.3. Table 5.3: Summary of parent, offspring and offspring mating combinations for the inbreeding problem, and related probabilities

Parents

Offspring

Offspring mating

AA × AA

All AA.

All AA × AA

AA × Aa

0.5 AA 0.5 Aa

0.25 AA × AA 0.5 AA × Aa 0.25 Aa × Aa

AA × aa

All Aa.

All Aa × Aa

Aa × Aa

0.25 AA 0.25 aa 0.5 Aa

0.0625 AA × AA 0. 25 AA × Aa 0.125 AA × aa 0.25 Aa × Aa 0.25 Aa × aa 0.0625 aa × aa

Aa × aa

0.5 Aa 0.5 aa

0.25 Aa × Aa 0.25 aa × aa 0.5 Aa × aa

aa × aa

All aa.

All aa × aa.

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Note that states 1 and 6 are absorbing states. R V S1 1 0 1 0 0 W 16 S 4 W S0 1 0 1 0 0 W S 2 W 4 S W 1 S0 0 0 8 0 0 W W. Hence the transition matrix will be P = S S0 1 1 1 1 0 W 4 4 W S 4 1 1 W S S0 0 0 4 2 0 W S 1 1 W SS0 0 0 16 4 1 WW T X P is not regular, since all powers of P will have zeroes in the first column except for the first position and in the last column except for the last position. What happens in the long term? We can easily use technology to find powers of the matrix P. To find the long-term steady-state solution if it exists we need to solve the homogeneous system (I – P)S = O, where I is the 6 × 6 identity matrix, S is the steady-state vector and O is the 6 × 1 zero matrix. This has been done with technology. The reduced row echelon form of (I – P) is R V S0 1 0 0 0 0 W S0 0 1 0 0 0 W S W S0 0 0 1 0 0 W S0 0 0 0 1 0 W S W S0 0 0 0 0 0 W S0 0 0 0 0 0 W T X and so, if S = [si] then s2 = s3 = s4 = s5 = 0, and all we know about s1 and R V S a W S 0 W S W 0 W S s6 is that they must sum to 1. In fact, for 0 ≤ a ≤ 1, S = is a vector S 0 W S W S 0 W S1 - a W T X such that PS = S, but it is not a steady-state vector.

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Using technology to compute large powers of P, we can guess that R V S1 0.75 0.5 0.5 0.25 0 W S0 0 0 0 0 0W S W 0 0 0 0 0 0W n S lim P = n"3 S0 0 0 0 0 0W S W 0 0 0 0W S0 0 S0 0.25 0.5 0.5 0.75 1 W T X This means that in the long term we will end up with some combination of AA × AA and/or aa × aa, depending on the initial state vector.

S t u d e n t ac t i v i t y 5 . 3 a

b

c

For Example 5.8, suppose that we always cross with dominants. Determine the transition matrix P, calculate P10 and P20, find lim P n and the steady-state vector if n"3 it exists. For Example 5.9, find the long-term distribution of the population if the initial state vector is R V R V R V S 0.1W S 0 W S 0 W S 0.2 W S 0. 1 W S0.25 W S 0.2 W S 0. 3 W S W W    iii  S0.35 W i  S W    ii  S S 0.2 W S 0.4 W S0.15 W S 0.2 W S 0.2 W S0.25 W S W S W S W S 0.1W S 0 W S 0 W T X T X T X There are three states in a country, called A, B and C. Each year 10% of the residents of state A move to state B and 30% to state C, 20% of the residents of State B move to state A and 20% to state C, and 5% of the residents of state C move to state A and 15% to state B. Suppose initially the population is equally divided between the three states. i Find the percentage of the population in the states after 3 years. ii Find the percentage of the population in the three states after a long period of time.

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SUMM A R Y

• A transition matrix P = [pij] for a Markov chain is a square matrix with non-negative entries such that the sum of the entries in each column is 1. • pij = probability of moving from state j to state i in one transition. • If the column vector S0 is the initial population distribution vector between states in a Markov chain with transition matrix P, the population distribution vector after one time period of the chain is PS0. • Pm is the transition matrix for m time periods, so the population distribution after m time periods is PmS0, and if pij(m) is the (i,j)th element of Pm, then pij(m) gives the probability of moving from state j to state i after m transitions. • A Markov chain and its associated matrix P is called regular if there exists an integer m such that Pm has no zero entries. • If P is a regular transition matrix for a Markov chain: – The columns of Pm all approach the same probability distribution vector S as m becomes large. – S is the unique probability vector satisfying PS = S. – As the number of time periods increases, the population distribution vectors approach S regardless of the initial population distribution (provided P is regular). Thus S is called the steadystate population distribution vector. 1-a b • For a 2 × 2 regular transition matrix = G, 0 ≤ a ≤ 1, a 1-b x b 0 ≤ b ≤ 1, the steady-state vector is = G, where x = and a+b y a y= (a + b > 0). a+b • A state is called an absorbing state if once the system is in this state it will never leave it. If state j is an absorbing state, then pjj = 1 and pij = 0 for i ≠ j References Anton, H & Rorres, C 2005, Elementary linear algebra (applications version), 9th edn, John Wiley and Sons, New York. Nicholson, KW 2001, Elementary linear algebra, 1st edn, McGraw-Hill Ryerson, Whitby, ON.

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chapter 5 Transition matrices Nicholson, KW 2003, Linear algebra with applications, 4th edn, McGraw-Hill Ryerson, Whitby, ON. Poole, D 2006, Linear algebra: A modern introduction, 2nd edn, Brooks Cole, California. Wheal, M 2003, Matrices: Mathematical models for organising and manipulating information, 2nd edn, Australian Association of Mathematics Teachers, Adelaide.

Websites http://math.ucsd.edu/~crypto/Monty/monty.html This website simulates the Monty Hall dilemma. http://orion.math.iastate.edu/msm/AthertonRMSMSS05.pdf#search=%22proof%20 that%20regular%20transition%20matrices%20converge%20to%20a%20steady %20state%22 – Iowa State Department of Mathematics This pdf discusses Markov chains. http://math.rice.edu/~pcmi/mathlinks/montyurl.html This website has links to many other sites that discuss or simulate the Monty Hall dilemma.

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6

C u rric u l u m c o n n e c t i o n s Different school systems and educational jurisdictions have particular features in their senior secondary mathematics curricula that have been developed over decades, and even centuries in some cases, to meet the historical and contemporary educational needs of their cultures and societies. When these curricula are reviewed, it is often the case that this includes a process of benchmarking with respect to corresponding curricula in other systems and jurisdictions. This may be in a local, county, state, national or international context. Over the past few decades, particularly in conjunction with renewed interest in comparative international assessments (such as TIMSS and PISA OECD), curriculum benchmarking has been employed extensively by educational authorities and ministries. Such benchmarking reveals much that is common in curriculum design and purpose in senior secondary mathematics courses around the world. Some key design constructs that are used to characterise the nature of senior secondary mathematics courses are: • content (areas of study, topics, strands) • aspects of working mathematically (concepts, skills and processes, numerical, graphical, analytical, problem-solving, modelling, investigative, computation and proof) • the use of technology, and when it is permitted, required or restricted (calculators, spreadsheets, statistical software, dynamic geometry software, CAS) • the nature of related assessments (examinations, school-based and the relationship between these) • the relationship between the final year subjects and previous years in terms of the acquisition of important mathematical background (assumed knowledge and skills, competencies, prerequisites and the like) • the amount and nature of prescribed material within the course (completely prescribed, unitised, modularised, core plus options)

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chapter 6 Curriculum connections

• the amount of in-class (prescribed) and out-of-class (advised) time that students are expected to spend for completion of the course In broad terms, it is possible to characterise four main sorts of senior secondary mathematics courses. Type 1 Courses designed to consolidate and develop the foundation and numeracy skills of students with respect to the practical application of mathematics in other areas of study. These often have a thematic basis for course implementation. Type 2 Courses designed to provide a general mathematical background for students proceeding to employment or further study with a numerical emphasis, and likely to draw strongly on data analysis and discrete mathematics. Such courses typically do not contain any calculus material, or only basic calculus material, related to the application of average and instantaneous rates of change. They may include, for example, business-related mathematics, linear programming, network theory, sequences, series and difference equations, practical applications of matrices and the like. Type 3 Courses designed to provide a sound foundation in function, coordinate geometry, algebra, calculus and possibly probability with an analytical emphasis. These courses develop mathematical content to support further studies in mathematics, the sciences and sometimes economics. Type 4 Courses designed to provide an advanced or specialist background in mathematics. These courses have a strong analytical emphasis and often incorporate a focus on mathematical proof. They typically include complex numbers, vectors, theoretical applications of matrices (for example transformations of the plane), higher level calculus (integration techniques, differential equations), kinematics and dynamics. In many cases Type 4 courses assume that students have previous or concurrent enrolment in a Type 3 course, or subsume them. Table 6.1 provides a mapping in terms of curriculum connections between the chapters of this book, the four types of course identified above, and the courses currently (2006) offered in various Australian states and territories.

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As this book is a teacher resource, these connections are with respect to the usefulness of material from the chapters in terms of mathematical background of relevance, rather than direct mapping to curriculum content or syllabuses in a particular state or territory. Table 6.1: Curriculum connections for senior secondary final year mathematics courses in Australia

State or territory

Type of course

Relevant chapters

Victoria

2: Further Mathematics

1, 2 and 5

3: Mathematical Methods (CAS)

all

4: Specialist Mathematics

1, 3 and 4

2: General Mathematics

3

3: Mathematics and Mathematics Extension 1

–

4: Mathematics Extension 2

–

2: Mathematics A

–

3: Mathematics B

3 and 4

4: Mathematics C

all

South Australia/ Northern Territory

2: Mathematical Applications

1, 2 and 5

3: Mathematical Methods/Mathematical Studies

all

4: Specialist Mathematics

–

Western Australia

2: Discrete Mathematics

–

3: Applicable Mathematics

1, 3 and 4

4: Calculus

–

2: Mathematics Applied

–

3: Mathematics Methods

–

4: Mathematics Specialised

1, 2, 3 and 4

New South Wales

Queensland

Tasmania

Table 6.2 provides a mapping in terms of curriculum connections between the chapters of this book, the four types of course identified above, and some of the courses currently (2006) offered in various English-speaking jurisdictions around the world. Again, as this book is a teacher resource, these

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chapter 6 Curriculum connections

connections indicate the usefulness of material from the chapters in terms of mathematical background of relevance, rather than direct mapping to curriculum content, or syllabuses, in a particular jurisdiction. Table 6.2: Curriculum connections for senior secondary final year mathematics courses in some jurisdictions around the world

State or territory

Type of course

Relevant chapters

College Board US

3: Advanced Placement Calculus AB

–

4: Advanced Placements Calculus BC

–

International Baccalaureate Organisation (IBO)

3: Mathematics SL

1, 2, 3 and 4

4: Mathematics HL

1, 2, 3 and 4

UK

3: AS Mathematics

1, 2, 3 and 4

4: Advanced level

1, 2, 3 and 4

Content from the chapters of the book may be mapped explicitly to topics within particular courses, and teachers will perhaps find it useful to informally make these more specific connections in terms of their intended implementation of a given course of interest to them. References The following are the website addresses of Australian state and territory curriculum and assessment authorities, boards and councils. These include various teacher reference and support materials for curriculum and assessment. The Senior Secondary Assessment Board of South Australia (SSABSA) http://www.ssabsa.sa.edu.au/ The Victorian Curriculum and Assessment Authority (VCAA) http://www.vcaa.vic.edu.au/ The Tasmanian Qualifications Authority (TQA) http://www.tqa.tas.gov.au/ The Queensland Studies Authority (QSA) http://www.qsa.qld.edu.au/ The Board of Studies New South Wales (BOS) http://www.boardofstudies.nsw.edu.au/ The Australian Capital Territory Board of Senior Secondary Studies (ACTBSSS) http://www.decs.act.gov.au/bsss/welcome.htm

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MathsWorks for Teachers Matrices The Curriculum Council Western Australia http://www.curriculum.wa.edu.au/ The following are the website addresses of various international and overseas curriculum and assessment authorities, boards, councils and organisations: College Board US Advanced Placement (AP) Calculus http://www.collegeboard.com/student/testing/ap/sub_calab.html?calcab International Baccalaureate Organisation (IBO) http://www.ibo.org/ibo/index.cfm Qualifications and Curriculum Authority (QCA) UK http://www.qca.org.uk/ OECD Program for International Student Assessment (PISA) http://www.pisa.oecd.org Trends in International Mathematics and Science Study (TIMSS) http://nces.ed.gov/timss/

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7

Solution notes to student ac t i v i t i e s

Student activity 2.1 R V 0.05W S R V R V S 4 0 1 3 2 2 W SS 0.1 WW S7.90 W S 5 1 0 0 4 2 W S 0.2 W S 8.35 W a S =S W# W S 0 0 0 4 3 0 W SS 0.5 WW S5.00 W S10 4 6 0 0 1 W S 1 W S 4.10 W T X S T X 2 W T X That is, Michael has $7.90, Jay has $8.35, Sam has $5.00 and Lin has $4.10. R V R V S7.90 W S6.004 W S 8.35 W S6.346 W b 0.76 # S W= S W S5.00 W S 3.80 W S 4.10 W S 3.116 W T X T X That is, Michael has US$6.00, Jay US$6.35, Sam US$3.80 and Lin US$3.12. Student activity 2.2 2 6 a i 2A = > H -2 4 ii iii iv

-1 4 A-B => H -3 -2 3 2 A+B == G 1 6 8 11 AB = = G 2 9

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v

3 4 BA = > H - 2 14

26 34 42 G 14 16 18 b Suppose MP = kP. Then we have two simultaneous equations 4p + q = kp 8p + 6q = kq and collecting terms in p and q, we have ]4 - kgp + q = 0 (i) 8p + ]6 - kgq = 0 (ii) Multiply equation (i) by 6 - k and subtract equation (ii) from it: ]6 - kg ]4 - kgp - 8p = 0 That is, ^16 - 10k + k2h p = 0. Since p ≠ 0, 16 - 10k + k2 = 0, so k = 2 or 8. vi

AC = =

c A2 = =

2 2 1 1 4 4 8 8 16 16 G = 2 = G, A 3 = = G, A 4 = = G, A 5 = = G 2 2 1 1 4 4 8 8 16 16

1 1 G = 2n - 1 A. 1 1 d Suppose A and O are of size n × n. Consider AO. The element in ith row Hence An = 2n - 1 =

n

n

k=1

k=1

and jth column of this product will be

/ aik okj = / aik 0 = 0, since

okj = 0 for all k = 1, 2 … n, j = 1, 2, 3 … n. Hence AO = O. The element in n

n

the ith row and jth column of OA is / oik akj = / 0akj = 0, and so k=1 k=1 OA = O. 1 1 2 1 Take A = = H, then AB = O, but neither A = O nor B = O. G, B = > -2 -2 6 3 -1 0 0 -1 0 -1 1 0 H#> H=> H =-= G =- I 0 -1 0 1 1 0 1 0 -1 0 -1 0 1 0 J 4 = J2 # J2 = > H#> H== G= I 0 -1 0 -1 0 1 f (X − Y)(X + Y) = X2 − YX + XY − Y2. Since generally YX ≠ XY for matrices, then (X − Y)(X + Y) ≠ X2 − Y2. 4 1 3 0 Take X = = G and Y = = G. 2 2 0 1

e J2 = >

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chapter 7 Solution notes to student activities

Then ]X - Y g ]X + Y g = =

1 1 7 1 9 4 G#= G== G and 2 1 2 3 16 5

4 12 3 02 18 6 9 0 9 6 G -= G == G-= G== G 2 2 0 1 12 6 0 1 12 5 and clearly ]X - Y g ]X + Y g ! X2 - Y2.

X2 - Y2 = =

4 0 3 G and Y = = 0 2 0 1 Then ]X - Y g ]X + Y g = = 0 Take X = =

0 G. 1 0 7 0 7 0 G#= G== G and 1 0 3 0 3

4 02 3 02 16 0 9 0 7 0 G -= G == G-= G== G 0 2 0 1 0 4 0 1 0 3 In this case ]X - Y g ]X + Y g = X2 - Y2.

X2 - Y2 = =

Student activity 2.3 1 4 -2 - 2> H -3 1 2 -3 ii > H -3 5 1 -1 -1 1 1 1 iii - 5 > H H = 5> 3 -2 -3 2 2 3 b This system can be written in matrix form as AX = B, where A = = G, 4 1 x 7 X = = G and B = = G. If A- 1 exists, then we can multiply both sides of the 3 y equation on the left by A- 1, and we thus have X = A- 1 B. Now, the inverse - 0.1 0.3 2 3 of A = = H. We can check this by matrix G is A- 1 = > 4 1 0.4 - 0.2 multiplication: - 0.2 + 1.2 0.6 - 0.6 2 3 - 0.1 0.3 1 0 H=> H== G> G = 4 1 0.4 - 0.2 - 0.4 + 0.4 1.2 - 0.2 0 1 Now, having found the inverse of the matrix A, we can solve the system of - 0.1 0.3 7 x 0.2 simultaneous linear equations: X = = G = A- 1 B = > H= G = = G y 0.4 - 0.2 3 2.2 Then x = 0.2 and y = 2.2 is the solution of the system of simultaneous linear equations given. a i

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Student activity 2.4 1 1 1 514229 317811 1 832040 G, f2 = = G, so f30 = A28 f2 = = G= G = = G. Hence 1 0 1 317811 196418 1 514229 the 29th number is 514 229 and the 30th is 832 040. Note that, in addition, the power of A also gives the 28th number 317 811 and 27th number 196 418.

A ==

Student activity 2.5 a Letter

S

A

V

E

space

T

H

E

Number

19

1

22

5

0

20

8

5

space

W

H

A

L

E

S

space

0

23

8

1

12

5

19

0

Letter Number

To code the message, we find the matrix product: 5 3 19 22 0 8 0 8 12 19 98 125 60 55 69 43 75 95 G= G== G 3 2 1 5 20 5 23 1 5 0 59 76 40 34 46 26 46 57 and so the message sent would be 98, 59, 125, 76, 60, 40, 55, 34, 69, 46, 43, 26, 75, 46, 95, 57. b To find the original message, we must first find the inverse of the coding matrix: 2 -3 5 3 -1 M- 1 = = H G => 3 2 -3 5 Arrange the received message in column vectors of length 2, and put them together into one matrix: 65 75 138 90 85 80 160 123 G = 42 50 87 54 54 49 99 76 Now multiply this on the left by M−1 to recover the original message as a matrix: 2 - 3 65 75 138 90 85 80 160 123 4 0 15 18 8 13 23 18 H= > G== G - 3 5 42 50 87 54 54 49 99 76 15 25 21 0 15 5 15 11 So the original message is represented by the numbers 4, 15, 0, 25, 15, 21, 18, 0, 8, 15, 13, 5, 23, 15, 18, 11 Checking the coding listed previously, we see that the message reads DO YOUR HOMEWORK.

=

144

chapter 7 Solution notes to student activities

Student activity 3.1 a {x + y = 1, 2x + y = −4} is one such system b In this case, one equation must be a multiple of the other. {2x + y = −4, 4x + 2y = −8} is one such system c Since (0, 0, 0) is a solution, the equations must be of the form ax + by + cz = 0. One example is {x − 2y + z = 0, 2x + 2y − 4z = 0} and the corresponding planes will intersect in a line with a one parameter family of solutions {(t, t, t): t ∈ R} Another example is {x − 2y + z = 0, 2x − 4y + 2z = 0} and the corresponding planes intersect in a plane (i.e. they are the same plane) with a two parameter family of solutions {(s, t, 2t − s): s, t ∈ R} d SOLVE([3·x – 2·y + z = 0, x – y – z = 10], [x, z]) 3y + 10 y + 30 x= / z =4 4 Solution set is 'c

3y + 10 y + 30 m : y ! R1 4 , y, - 4

Student activity 3.2 a It is clear that (0, 0) is a solution of this system of equations. The a b coefficient matrix is = G. This is invertible if ad − bc ≠ 0, in which case c d x a b -1 0 0 G = G = = G, the unique solution. = G== y 0 0 c d So we need to consider the case ad = bc. If c and d are both non-zero, a b then c = and the equations are a multiple of one another, and so there d will be infinitely many solutions. If, say, c is zero, then either a or d is zero. If d is zero, then the second equation is 0x + 0y = 0, which has R2 as solution, and so the solution to the system is simply the set of points on the line ax + by = 0. If a is zero, then the equations are {by = 0, dy = 0}, so one is a multiple of the other, and the solution set is the set of points on the x-axis (provided one of b or d is non-zero). (A similar argument applies if d = 0.) b The system of linear equations {ax + by = e, cx + dy = f} will have infinitely many solutions when one equation is simply a non-zero multiple a b e of the other. Assuming c ≠ 0, d ≠ 0 and f ≠ 0, then we need c = = . f d

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This can be written ad − bc = 0 and af − ce = 0 and bf − de = 0, and this way we do not have to assume c ≠ 0, d ≠ 0 and f ≠ 0. The system of linear equations {ax + by = e, cx + dy = f} will have no solutions if the corresponding lines are parallel but distinct. In this case we require ad − bc = 0 but either af − ce ≠ 0 or bf − de ≠ 0. The system of linear equations will have a unique solution if ad − bc ≠ 0. c We have x = 3t + 1 and y = 2t − 1. We need to eliminate t from these x-1 equations. From the first equation, t = 3 . Use this to substitute for t in the second equation: x-1 y = 2 b 3 l - 1 or 3y = 2x - 5 d The underlying equation is 3x − y = 4, or y = 3x – 4. So simply choose any expression involving a parameter for x, and use the equation to write y as a function of the parameter. Equivalent solution sets would be {(r + 1, 3r – 1): r ∈ R} and {(s − 1, 3s – 7): s ∈ R}. Student activity 3.3

R V S1 2 - 1 2 W a The augmented matrix is S1 4 - 3 3 W, which has reduced row echelon form S W R V S2 5 - 3 1 W S1 0 1 0 W T X S0 1 - 1 0 W. Looking at the last row of the reduced form (corresponding to SS W 0 0 0 1W T X the equation 0x + 0y + 0z = 1) we see the equations are inconsistent. Hence there is no solution. R V S- 1 1 1 - 1 W b The augmented matrix is S 1 - 1 3 5 W, which has reduced row echelon S W S 3 - 2 1 - 2W Zx =- 7_ R V T X S1 0 0 - 7 W ] b form S0 1 0 - 9 W. This corresponds to the system of equations [ y =- 9` S W ] b S0 0 1 1 W z=1 a \ T X and hence we have a unique solution (−7, −9, 1). R V S 2 2 -1 5 W c The augmented matrix is S- 2 1 1 7 W, which has reduced row echelon S W S- 4 1 2 10 W R V T X S1 0 - 1 - 3 W 2 2W S form S0 1 0 4 W. There are two leading variables (corresponding to the SS0 0 0 0 WW T X

146

chapter 7 Solution notes to student activities

two leading 1s) x and y, and one free variable z. So let z = k, an arbitrary k-3 real number. Then the solution set is 'b 2 , 4, kl: k ! R1. Student activity 3.4 a Let v, x, y and z be the unknown scores of players 1, 2, 3 and 4 respectively, and a, b, c and d the known totals. Then v+x = a

x+y = b y+z = c z+v = d is a system of four equations in four unknowns. The augmented matrix is R V R V S1 1 0 0 a W S1 0 0 1 0 W S0 1 1 0 b W S0 1 0 - 1 0 W S W and its reduced row echelon form using CAS is S W. S0 0 1 1 c W S0 0 1 1 0 W S1 0 0 1 d W S0 0 0 0 1 W T X T X Now we have a problem. There is no sign of a, b, c or d, and the last line corresponds to an equation which reads





0v + 0x + 0y + 0z = 1 which is clearly a contradiction, indicating there are no solutions. So to find out what is really happening, we need to use the Gaussian elimination procedure to reduce the augmented matrix to echelon form (but not reduced). To begin, the new row 4 will be the old row 4 subtract row 1, giving R V S1 1 0 0 a W S0 1 1 0 b W S W. S0 0 1 1 c W S0 - 1 0 1 d - a W T X Next, the new row 4 should be the old row 4 with row 2 added to it. R V a S1 1 0 0 W S0 1 1 0 W b S W c S0 0 1 1 W S0 0 1 1 d - a + b W T X

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MathsWorks for Teachers Matrices

Next, the new row 4 will be the old row 4 with row 3 subtracted from it. R V a S1 1 0 0 W S0 1 1 0 W b S W c S0 0 1 1 W S0 0 0 0 d - a + b - c W T X Now the last equation is 0v + 0x + 0y + 0z = d − a + b − c. Generally, if a + c ≠ b + d, there will be no solution. In this example, (v + x) + (y + z) = a + c, and (x + y) + (z + v) = b + d, so a + c = b + d = sum of all players scores in the tournament, and this means R V S1 1 0 0 a W S0 1 1 0 b W that the echelon form of the augmented matrix is S W, which is S0 0 1 1 c W S0 0 0 0 0 W X R T V S1 0 0 1 a - b + c W S0 1 0 - 1 b - c W equivalent to the reduced row echelon matrix S W. c S0 0 1 1 W S0 0 0 0 W 0 T X So there will be three basic variables v, x, and y, and one free variable z, and hence there will be an infinite number of solutions and thus the scores will not be able to be uniquely determined. Writing z = k, k ∈ R, the first row of the matrix tells us that v = −k + a − b + c, the second row that x = k + b − c, and the third row that y = −k + c. b Let f(x) = ax3 + bx 2 + cx + d be the equation of a cubic polynomial function, with a, b, c and d the unknown coefficients. From Example 3.15 we have three equations: a+b+c+d = 0 -a + b - c + d = 0 3a + 2b + c =- 4 Since f ′(x) = 3ax 2 + 2bx + c, and the slope at x = −1 is 12, so 3a − 2b + c = 12 is a fourth equation to add to the system of three above.

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chapter 7 Solution notes to student activities





The augmented matrix for this system of four equations is R V S 1 1 1 1 0 W S- 1 1 - 1 1 0 W S W S 3 2 1 0 - 4W S 3 - 2 1 0 12 W T X which reduces to R V S1 0 0 0 2 W S0 1 0 0 - 4 W S W S0 0 1 0 - 2 W S0 0 0 1 4 W T X and so a = 2, b = −4, c = −2 and d = 4, giving f(x) = 2x3 − 4x 2 − 2x + 4 The graph of this function is given in the figure below. y 4 3 2 1 –4

–3

–2

–1

O –1

1

2

3

x

–2 –3

c Let f(x) = ax4 + bx3 + cx 2 + dx + e be the rule for the family of quartic polynomials. Since they pass through (1, 2) and (−2, −1) and have slope 5 at x = 1, we have the following equations (see Example 3.16 for details): a+b+c+d+e = 2 16a - 8b + 4c - 2d + e =- 1 4a + 3b + 2c + d = 5 In addition, they must pass through the point (2, 0), so f(2) = 0 and 16a + 8b + 4c + 2d + e = 0.

149

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This gives us a system of 4 equations in 5 unknowns. The augmented R V S1 0 0 0 1 - 35 W 4 24 W R V S S1 1 1 1 1 2 W S0 1 0 0 - 1 5 W S16 - 8 4 - 2 1 - 1W S 2 6 W matrix is S . W, which reduces to S 3 137 W S4 3 2 1 0 5 W S0 0 1 0 - 4 24 W S16 8 4 2 1 0 W S W T X S0 0 0 1 2 - 37 W S 12 W T X Now there are four leading variables a, b, c and d, and one free variable e. Let e = t, t ∈ R. Then the first row of the reduced matrix gives the equation 1 35 a + 4 e =- 24 1 35 and so a =- 4 t - 24 , since e = t. The second row of the reduced matrix gives the equation 1 5 b - 2e = 6 1 5 and so b = 2 t + 6 . The third row of the reduced matrix gives the equation 3 137 c - 4 e = 24 3 137 and so c = 4 t + 24 . The fourth row of the reduced matrix gives the equation 37 d + 2e =- 12 37 and so d =- 2t - 12 .

Hence the family of functions is of the form: 1 35 1 5 3 137 37 f]xg = b- 4 t - 24 l x 4 + b 2 t + 6 l x 3 + b 4 t + 24 l x2 + b- 2t - 12 l x + t, t d R Again, we will plot some members of this family. 35 5 137 37 When t = 0, f]xg = b- 24 l x 4 + b 6 l x 3 + b 24 l x2 + b- 12 l x 41 4 155 61 When t = 1, f]xg = b- 24 l x 4 + b 3 l x 3 + b 24 l x2 + b- 12 l x + 1 13 19 7 155 When t =- 8, f]xg = b 24 l x 4 + b- 6 l x 3 + b- 24 l x2 + b 12 l x - 8 The graphs of these are given in the figure on the following page.

150

chapter 7 Solution notes to student activities y 20 15 10 5 –10

–8

–6

–4

–2

O –5

2

4

6

8

10

x

–10 –15 –20 –25 –30

d Let f(x) = ax4 + bx3 + cx 2 + dx + e be the rule for the family of quartic polynomials. Since they pass through (1, 2), (−2, −1) and (2, 0), and have slope 5 at x = 1, we have the following equations (see example above for details): a + b + c + d + e =2 16a - 8b + 4c - 2d + e = - 1 4a + 3b + 2c + d =5 16a + 8b + 4c + 2d + e = 0 Now we are looking for the particular curve that passes through (−1, 5), so f(−1) = 5, and this gives a fifth equation a − b + c − d + e = 5. The augmented matrix for this system of 5 equations is R V S1 0 0 0 0 - 4 W 3W S R V 1 1 1 1 1 2 7 S0 1 0 0 0 W S W S 12 W S16 - 8 4 - 2 1 - 1W S S W 16 W S 4 3 2 1 0 5 W, which has reduced echelon form S0 0 1 0 0 3 W. S W S16 8 4 2 1 0 W S0 0 0 1 0 - 25 W S W 12 W S 1 -1 1 -1 1 5 W S T X S 1W S0 0 0 0 1 - 2 W T X 4 7 16 25 1 Hence f]xg =- 3 x 4 + 12 x 3 + 3 x2 - 12 x - 2 , and its graph is given in the figure on the following page.

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MathsWorks for Teachers Matrices y 8 6 4 2 –5

–4

–3

–2

–1

O –2

1

2

3

4

x

–4 –6 –8 –10

Student activity 4.1 a b 0 0 a = G = G = = G and hence any linear transformation maps the origin to 0 c d 0 itself. 3 -3 b T cannot be uniquely determined since > H =- 1 > H, that is, one vector -1 1 is a multiple of the other. a b Let = G be the matrix for T. c d -1 1 a b 3 a b -3 G > H = > H and = G > H = > H. -1 c d -1 c d 1 1 3a - b =- 1 - 3a + b = 1 This yields the equations and . 3c - d = 1 - 3c + d =- 1 Grouping the equations for a and b and for c and d together, we have 3a - b =- 1 3c - d = 1 ) 3 and ) 3 - 3a + b = 1 - 3c + d =- 1 But the two equations in each set are in fact a multiple of one another, so we only have two distinct equations in 4 unknowns. If we choose any values for b and d, then the values of a and c will be determined by these equations. 1 4 Take b = 4 and d = −1. Then a = 1 and c = 0, giving > H as a matrix 0 -1 for T. Then =



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chapter 7 Solution notes to student activities

1 4 Take b = 4 and d = 2. Then a = 1 and c = 1, giving = G as a matrix 1 2 for T. 1 4 Take b = 4 and d = −4. Then a = 1 and c = −1, giving > H as a -1 -4 matrix for T, (and this matrix is singular). 4 3 x 1 4 3 r 0 c Let = G = G = = G and = G = G = = G. 5 4 y 0 5 4 s 1 x r 4 3 -1 1 4 3 -1 0 Then = G = = G = G and = G = = G = G. y 5 4 0 s 5 4 1 4 -3 1 4 -3 0 4 -3 x r Then = G = > H = G = > H and = G = > H = G = > H. -5 y -5 4 0 s -5 4 1 4 So (4, −5) is mapped to (1, 0) and (−3, 4) is mapped to (0, 1). Student activity 4.2 a b 0 0 G = G = = G and hence any linear transformation maps the 0 c d 0 origin to itself. Property 2: A non-singular linear transformation is onto, since if A is the matrix of the transformation then any point v = (x, y) is the image of A−1v. A non-singular linear transformation is one-to-one, since if A is the matrix of the transformation, then if Av1 = Av2 for any v1, v2 ∈ R2, then A(v1 − v2) = O, so A−1 A(v1 − v2) = A−1O, that is, (v1 − v2) = O, and so v1 = v2. Property 3: Consider a straight line with vector equation r = u + tv, and a transformation T with matrix A. Then T(r) = Au + tAv = u1 + tv1, where u1, tv1, ∈ R2, and so is the vector equation for a line. Property 4: Two distinct parallel lines will have vector equations of the form r1 = u1 + tv and r2 = u2 + tv, where u1 ≠ u2 and v gives the direction of the lines. Under the transformation with matrix A, the direction of both lines will be Av, so will still be parallel and distinct, since Au1 ≠ Au2. Property 5: A line through the origin can be written in vector form r = tv, and if A is the matrix of the transformation then T(r) = tAv, which also passes through the origin.

a Property 1: =

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MathsWorks for Teachers Matrices

x1 2 1 x b > H = = G= G y1 1 3 y x 2 1 - 1 x1 G > H = G== y1 y 1 3 =>

3 -1 5 5 -1 2 5 5

1 H >y H

x

1

x = 53 x1 - 15 y1 y=

-1 2 5 x1 + 5 y1

y = 5 - 3x becomes

- 15 x1 + 25 y1 = 5 - 3` 53 x1 - 15 y1j

which simplifies to y1 = 8x1 - 25 or y = 8x - 25

x1 1 1 x c > H = = G = G y1 1 1 y x1 = x + y y1 = x + y ` Since x + y ! constant, x1 = y1 so y = x is image of y = 5 - 3x. d If the line maps to a point, then from c x1 = y1 = k, where k is a real constant. So lines of form x + y = k are mapped to the point (k, k) under this transformation. 2 1 0 0 e = G= G = = G 1 3 0 0 2 1 0 1 G= G = = G = 1 3 1 3 2 1 1 3 G= G = = G = 1 3 1 4 2 1 1 2 G= G = = G = 1 3 0 1 So the transformation maps the unit square to the parallelogram with vertices (0, 0), (1, 3), (3, 4) and (2, 1), and the area of the parallelogram is 2 1 det = G × area of unit square = 5 square units. 1 3

154

chapter 7 Solution notes to student activities

Student activity 4.3 a =

x1 3 0 x G= G = > H y1 0 2 y

R V S x1 W x 3 1 2 0 x1 G > H = SS WW = G = 6= y y y 0 3 1 1 S2W T X y x x Hence the image of y = sin(x) is 21 = sin c 31 m, or y = 2 sin b 3 l. x1 3 5 x b = G= G = > H y 1 2 y 1 2x1 - 5y1 2 - 5 x1 x H> H = > H = G=> - x1 + 3y1 y - 1 3 y1 Hence the image of y = x 2 is −x1 + 3y1 = (2x1 − 5y1)2, or 4x 2 − 20xy + 25y2 + x − 3y = 0. x1 1 1 x G= G = > H y1 1 1 y So x + y = x1 = y1, and so the image of y = x 2 is y = x.

c =

Student activity 4.4 a >

ky 0 0 0 H= G = = G 0 0 kx 0

>

ky 0 1 ky H= G = > H 0 kx 0 0

>

ky 0 1 ky H= G = > H 0 kx 1 kx

>

ky 0 0 0 H = G = >k H 0 kx 1 x

Hence the area of the rectangle formed by the transformed vertices is kxky. ky 0 x x1 b > H = G = >y H 0 kx y 1 So x =

y x x1 y and y = 1 , and so y = f(x) is transformed to 1 = f e 1 o, that is, kx ky ky kx

y = kx f e

x . ky o

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MathsWorks for Teachers Matrices

x y c From part b, x = a1 and y = 1 , and so x 2 + y2 = 1 is transformed to b 2 2 2 2 c x1 m + c y1 m = 1, that is b x l + c y m = 1. This is the equation of an ellipse a a b b with horizontal axis length 2a and vertical axis length 2b.

a 0 G = πab. 0 b d The matrix for an anticlockwise rotation about the origin through an angle cos ]qg - sin ]qg of θ is = G. sin ]qg cos ]qg The matrix for an anticlockwise rotation about the origin through an cos ^fh - sin ^fh angle of φ is > H, and the matrix for an anticlockwise rotation sin ^fh cos ^fh cos ^q + fh - sin ^q + fh about the origin through an angle of θ + φ is > H. sin ^q + fh cos ^q + fh Hence area of ellipse = area of unit circle × det =

Then, since a rotation through an angle of θ followed by a rotation through an angle φ is equivalent to a rotation through an angle θ + φ:



=

cos ^q + fh - sin ^q + fh cos ]qg - sin ]qg cos ^fh - sin ^fh H=> H G> sin ]qg cos ]qg sin ^fh cos ^fh sin ^q + fh cos ^q + fh

cos ]qg - sin ]qg cos ^fh - sin ^fh H= G> sin ]qg cos ]qg sin ^fh cos ^fh cos ]qg cos ^fh - sin ]qg sin ^fh - cos ]qg sin ^fh - sin ]qg cos ^fh > H sin ]qg cos ^fh + cos ]qg sin ^fh cos ]qg cos ^fh - sin ]qg sin ^fh And so cos ^q + fh = cos ]qg cos ^fh - sin ]qg sin ^fh and sin ^q + fh = sin ]qg cos ^fh + cos ]qg sin ^fh. x + 3y 1 3 x e = H so the points (0, 0), (1, 0), (1, 1) and (0, 1) are G= G = > 0 1 y y transformed to the points (0, 0), (1, 0), (4, 1) and (3, 1) respectively. The unit square and its image under this transformation are shown on the following page. Now =

156

chapter 7 Solution notes to student activities y 2

1

–1

O

1

2

3

4

x

–1

f =

x1 1 3 x G= G = > H y1 0 1 y

x1 - 3y1 1 - 3 x1 x 1 3 - 1 x1 so = G = = H> H = > H. G > H=> y1 y1 y 0 1 0 1 y1 Hence the function y = f(x) is transformed to y = f(x − 3y) under this transformation. The image of y = x 2 will be y = (x − 3y)2.

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MathsWorks for Teachers Matrices

Student activity 4.5

a Dilation by factor 0.5 from y-axis has matrix =

0.5 0 G 0 1

-1 0 Reflection in y-axis has matrix > H 0 1 Translation 3 units parallel to x-axis and 1 unit parallel to y-axis is 3 represented by = G 1 x1 - 1 0 0.5 0 x 3 Hence > H= G= G + = G = > H y1 1 0 1 0 1 y - 2^x1 - 3h x 0.5 0 - 1 - 1 0 - 1 x1 - 3 Then = G = = H > H=> G > H y 1 y1 - 1 y 0 1 0 1 1

1 So y = f(x) becomes y – 1 = f(−2(x – 3)) and hence y = x is transformed 1 1 + 1. to y - 1 = or y = - 2]x - 3g - 2]x - 3g 1 0 2 H gives reflection in the x-axis, > H gives a translation of 2 units to 0 -1 -1 the right and 1 unit down. x1 1 0 x 2 Then > H = > H= G + > H y1 0 -1 y -1

b >

x1 2 1 0 x and > H - > H = > H= G y1 -1 0 -1 y that is, >

x1 - 2 1 0 x H = G, and so H=> y1 + 1 0 -1 y

x1 - 2 1 0 - 1 x1 - 2 1 0 x1 - 2 x H > H> H=> H=> H = G=> y + 1 y + 1 0 -1 0 -1 1 ^y1 + 1h y 1 So x − y2 = 0 is transformed to (x − 2) − (y + 1)2 = 0.

158

chapter 7 Solution notes to student activities

Student activity 5.1 a S1 = PS0 = =

0 0.25 0.5 0.125 G= G = = G 1 0.75 0.5 0.875

S2 = P2 S0 = =

0 0.25 2 0.5 0.21875 G = G== G 1 0.75 0.5 0.78125

S5 = P5 S0 = =

0 0.25 5 0.5 0.19971 G = G.= G 1 0.75 0.5 0.80029

S10 = P10 S0 = =

0 0.25 10 0.5 0.20000 G = G.= G 1 0.75 0.5 0.80000

0 0.25 50 0.5 0.20000 G = G.= G 1 0.75 0.5 0.80000 0.445 0.444 b P 3 = = G 0.555 0.556 Hence the probability of going from state 1 to state 2 in 3 transitions is the element in (2, 1) position of P3, which is 0.555. Similarly, the probability of going from state 2 to state 2 in 3 transitions is the element in (2, 2) position of P3, which is 0.556. S 50 = P 50 S0 = =

Student activity 5.2 a P2 = =

0.75 0.50 G has no non-zero entries, so P is regular. 0.25 0.50 2 3

2 3

3

3

Since a = 12 and b = 1, S = > 1 H and L = > 1

2 3 1 3

H.

0 1 G = P3 = P5 = f 1 0 1 0 P2 = = G = I = P 4 = P6 = f 0 1 P is not regular. The limit matrix L does not exist, as powers of P 0 1 1 0 oscillate between = G and = G. There is no steady-state vector; however, 1 0 0 1 0.5 if S = = G, then PS = S. 0.5 c It is not regular since any power of P will have a zero in (1, 2) position. (In

b P = =

fact, Pn = >

0 0 0 an 0 H.)The steady-state vector is S = = G and lim Pn = = G. n n"3 1 1 1 1-a 1

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Student activity 5.3 R V S1 1 0 W S 2 W 1 a P = SS0 2 1 WW S0 0 0 W TR X V S1 1023 511 W S 1024 512 W 1 1 P10 = SS0 1024 512 WW S0 0 0 W TR X V S1 1048575 524287 W S 1048576 524288 W 1 1 20 P = SS0 1048576 524288 WW S0 0 0 W T R X V S1 1 1 W lim Pn = S0 0 0 W n"3 SS W 0 0 0W T X R V S1 W Steady-state vector S = S 0 W SS WW 0 T X R V J N 1 W S 1 2 0 R V R V K O S WS 1 W S 1 W K O K Check: PS = S0 1 1 WS 0 W = S 0 W = SO S 2 WS W S W K O K O S0 0 0 W S 0 W S 0 W T X T X L T X R VR V P R V S1 0.75 0.5 0.5 0.25 0 WS0.1 W S0.5W S0 0 0 0 0 0 WS0.2 W S 0 W S WS W S W 0 0 0 0 0 0 WS0.2 W S 0 W S b i = S0 0 0 0 0 0 WS0.2 W S 0 W S WS W S W 0 0 0 0 WS0.2 W S 0 W S0 0 S0 0.25 0.5 0.5 0.75 1 WS0.1 W S0.5W T X T X TR XV R V R V S 19 W 1 0 . 75 0 . 5 0 . 5 0 . 25 0 S WS 0 W S 40 W S0 0 0 0 0 0 WS 0.1 W S 0 W S WS W S W 0 0 0 0 0 0 WS 0.3 W S 0 W ii S = S0 0 0 0 0 0 WS0.4 W S 0 W S WS W S W 0 0 0 0 WS0.2 W S 0 W S0 0 S0 0.25 0.5 0.5 0.75 1 WS 0 W S 21 W T X T X S 40 W T X

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chapter 7 Solution notes to student activities

iii

c

R S1 0.75 S0 0 S S0 0 S0 0 S S0 0 S0 0.25 T

0.5 0 0 0 0 0.5

0.5 0.25 0 0 0 0 0 0 0 0 0.5 0.75

VR V R V 0 WS 0 W S0.5W 0 WS0.25W S 0 W WS W S W 0 WS0.35W S 0 W = 0 WS0.15W S 0 W WS W S W 0 WS0.25W S 0 W 1 WS 0 W S0.5W XT X T X Current state

State next year

A

B

C

A

0.6

0.2

0.05

B

0.1

0.6

0.15

C

0.3

0.2

0.8

R V S0.6 0.2 0.05W Then the transition matrix is P = S0.1 0.6 0.15W, and the initial state R V SS W 0.3 0.2 0.8 W S1W T X S3W 1W S vector is S0 = S 3 W, and so the distribution after 3 years is S1W SS 3 WW T X R V R V3 S 13 W R V S0.6 0.2 0.05W S 1 W S0.226 W P 3 S0 = S0.1 0.6 0.15W S 3 W . S0.256 W SS W S W 0.3 0.2 0.8 W SS 1 WW S0.518 W T X S3W T X T X After 3 years approximately 22.6% of the population will live in state A, 25.6% in state B and 51.8% in state C. To find the long term population distribution, we can investigate powers of P. R V S0.1961 0.1961 0.1961 W P15 . S0.2549 0.2549 0.2549 W SS W 0.5490 0.5490 0.5490 W TR XV S 0.1961 0.1961 0.1961 W P20 . S0.2549 0.2549 0.2549 W SS W 0.5490 0.5490 0.5490 W T X

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MathsWorks for Teachers Matrices

It appears that, in the long term, approximately 19.6% of the population will live in state A, 25.5% in state B and 54.9% in state C. Alternatively, consider V R V R V R S1 0 0 W S0.6 0.2 0.05W S 0.4 - 0.2 - 0.05W I - P = S0 1 0 W- S0.1 0.6 0.15W = S- 0.1 0.4 - 0.15W W SS W S W S 0 0 1 W S0.3 0.2 0.8 W S- 0.3 - 0.2 0.2 W T X T X R X T V S1 0 - 5 W R V 14 W S Sx W - 13 which has reduced echelon form SS0 1 28 WW, and if S = Sy W, then we have SS WW S0 0 0 W z T X T X 5z 13z x = 14 , y = 28 and x + y + z = 1, which has solution 10 13 28 x = 51 / y = 51 / z = 51 , or, approximately, x = 0.1961, y = 0.2549, z = 0.5490.

162

R e f e r e n c e s a n d f u r t h e r reading Adler, I 1958, The new mathematics, John Day, New York. Allendoerfer, CB & Oakley, CO 1955, Principles of mathematics, McGraw-Hill, New York. Anton, H & Rorres, C 2005, Elementary linear algebra (applications version), 9th edn, John Wiley and Sons, New York. Cameron, N, Clements, K, Green, LJ & Smith, GC 1970, School Mathematics Research Foundation: Pure mathematics, Chesire, Melbourne. Cirrito, F (ed.) 1999, Mathematics higher level (core), 2nd edn, IBID Press, Victoria. Evans, L 2006, Complex numbers and vectors, ACER Press, Camberwell. Fitzpatrick, JB & Galbraith, P 1971, Pure mathematics, Jacaranda, Milton, Queensland. Fraleigh, JB & Beauregard, AR 1995, Linear algebra (3rd edition with historical notes by VJ Katz), Addison-Wesley, Reading, MA. Garner, S, McNamara, A & Moya, F 2004, CAS analysis supplement for Mathematical Methods CAS Units 3 and 4, Pearson, Melbourne. Grattan-Guinness, I (ed.) 1994, Companion encyclopedia of the history and philosophy of the mathematical sciences, volume 1, Routledge, London. Hill, RO Jr 1996, Elementary linear algebra with applications, 3rd edn, Saunders College, Philadelphia. Hodgson, B & Patterson, J 1990, Change and approximation, Jacaranda, Melbourne. Hoe, J 1980, ‘Zhu Shijie and his jade mirror of the four unknowns’, in JN Crossley (ed.), First Australian conference on the history of mathematics: Proceedings, Monash University, Clayton. Holton, DA & Pye, W 1976, Aggregating algebra, Holt, Rhinehart and Winston, Sydney. Kangshen, S, Crossley, JN & Lun, A 1999, The nine chapters on the mathematical art: Companion and commentary, Oxford University Press, Oxford. Katz, VJ 1993, A history of mathematics—an introduction, Harper-Collins, Reading, MA. Kemeny, JG, Scheifer, A, Snell, JL & Thompson, GC 1972, Finite mathematics with business applications, Prentice-Hall, Englewood Cliffs, NJ. Kissane, B 1997, More mathematics with a graphics calculator, Mathematical Association of Western Australia, Claremont. Leigh-Lancaster, D 2006, Functional equations, ACER Press, Camberwell. Lipschutz, S & Lipson, M 2000, Schaum’s outline of linear algebra, 3rd edn, McGraw-Hill, New York. Nicholson, WK 2001, Elementary linear algebra, 1st edn, McGraw-Hill Ryerson, Whitby, ON.

163

MathsWorks for Teachers Matrices Nicholson, WK 2003, Linear algebra with applications, 4th edn, McGraw-Hill Ryerson, Whitby, ON. Poole, D 2006, Linear algebra: A modern introduction, 2nd edn, Brooks Cole, California. Sadler, AJ & Thorning, DWS 1996, Understanding pure mathematics, Oxford University Press, Oxford. Victorian Curriculum and Assessment Authority (VCAA) 2005, VCE Mathematics study design, VCAA, East Melbourne. Wheal, M 2003, Matrices: Mathematical models for organising and manipulating information, 2nd edn, Australian Association of Mathematics Teachers, Adelaide.

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Notes

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MathsWorks for Teachers

MathsWorks for Teachers Series editor David Leigh-Lancaster

This book provides mathematics teachers with an elementary introduction to matrix algebra and its uses in formulating and solving practical problems, solving systems of linear equations, representing combinations of affine (including linear) transformations of the plane and modelling finite state Markov chains. The basic theory in each of these areas is explained and illustrated using a broad range of examples. A feature of the book is the complementary use of technology, particularly computer algebra systems, to do the calculations involving matrices required for the applications. A selection of student activities with solutions and text and web references are included throughout the book.

Pam Norton

Matrices are used in many areas of mathematics, and have also have applications applications in diverse in diverse areas such areas as such engineering, as engineering, computer computer graphics,graphics, image processing, image processing, physicalphysical sciences, sciences, biological biological sciences and sciences socialand sciences. social Powerful sciences. Powerful calculators calculators and computers and computers can now can now carry outcarry complicated out complicated and difficult and numeric difficult numeric and algebraic and algebraic computations computations involving involving matrix methods, matrix and methods, such technology and such technology is a vital tool is ain vital related tool real-life, in relatedproblemreal-life, problem-solving solving applications. applications.

MATRICES

Matrices Pam Norton

Series overview

MathsWorks for Teachers has been developed to provide a coherent and contemporary framework for conceptualising and implementing aspects of middle and senior mathematics curricula.

Matrices

Titles in the series are: Functional Equations David Leigh-Lancaster Contemporary Calculus Michael Evans Matrices Pam Norton Foundation Numeracy in Context David Tout & Gary Motteram Data Analysis Applications Kay Lipson Complex Numbers and Vectors Les Evans

Functional Equations

Contemporary Calculus

David Leigh-Lancaster

Michael Evans

MathsWorks for Teachers

MathsWorks for Teachers

Matrices Pam Norton

MathsWorks for Teachers

Foundation Numeracy in Context

Data Analysis Applications

Complex Numbers and Vectors

David Tout and Gary Motteram

Kay Lipson

Les Evans

MathsWorks for Teachers

MathsWorks for Teachers

MathsWorks for Teachers

Pam Norton

ISBN 978-0-86431-508-3

9

780864 315083

MathsWorks for Teachers