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Engineering Mathematics-1 About the Author T Veerarajan is the retired Dean, Department of Mathematics, Velammal College of Engineering and Technology, Viraganoor, Madurai, Tamil Nadu. A Gold Medalist from Madras University, he has had a brilliant academic career all through. He has 50 years of teaching experience at undergraduate and postgraduate levels in various established Engineering Colleges in Tamil Nadu including Anna University, Chennai. Engineering Mathematics T Veerarajan Dean, Department of Mathematics, (Retd.) Velammal College of Engineering & Technology Viraganoor, Madurai, Tamil Nadu McGraw Hill Education (India) Private Limited NEW DELHI McGraw Hill Education Ofﬁces New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited P-24, Green Park Extension, New Delhi 110 016 Engineering Mathematics-1 Copyright © 2014, by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listing (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. ISBN (13 digits): 978-93-5134-174-1 ISBN (10 digits): 93-5134-174-7 Vice President and Managing Director: Ajay Shukla Head—Higher Education Publishing and Marketing: Vibha Mahajan Publishing Manager—SEM & Tech. Ed: Shalini Jha Editorial Executive—Acquisitions: Vamsi Deepak Sankar Executive—Editorial Services: Sohini Mukherjee Manager—Production Systems: Satinder S Baveja Production Executive: Anuj K Shriwastava Marketing Manager—Higher Education: Vijay Sarathi Product Specialist—SEM & Tech Ed: Sachin Tripathi Sr Graphic Designer (Cover): Meenu Raghav General Manager—Production: Rajender P Ghansela Production Manager—Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Text-o-Graphics, B1/56, Arawali Apartment, Sector 34, Noida 201301 and Cover Printer: Contents Preface xi 1. Matrices I-1.3 1.1 Introduction I-1.3 1.2 Vectors I-1.4 1.3 Linear Dependence and Linear Independence of Vectors I-1.5 1.4 Methods of Testing Linear Dependence or Independence of a Set of Vectors I-1.5 1.5 Consistency of a System of Linear Algebraic Equations I-1.5 Worked Example 1(a) I-1.7 Exercise 1(a) I-1.24 1.6 Eigenvalues and Eigenvectors I-1.27 Worked Example 1(b) I-1.32 Exercise 1(b) I-1.44 1.7 Cayley – Hamilton Theorem I-1.46 1.8 Property I-1.47 1.9 Calculation of Powers of a Matrix A I-1.49 1.10 Diagonalisation by Orthogonal Transformation or Orthogonal Reduction I-1.49 Worked Example 1(c) I-1.49 Exercise 1(c) I-1.62 1.11 Quadratic Forms I-1.64 Worked Example 1(d) I-1.67 Exercise 1(d) I-1.74 Answers I-1.75 2. Sequences and Series I-2.1 I-2.1 2.2 General Properties of Series I-2.3 Worked Example 2(a) I-2.8 Exercise 2(a) I-2.21 Worked Example 2(b) I-2.23 Exercise 2(b) 1-2.36 2.3 Alternating Series I-2.38 Worked Example 2(c) I-2.42 Exercise 2(c) I-2.47 vi Contents 2.4 Sequences and Series I-2.49 Exercise 2(d) I-2.50 Worked Example 2(d) I-2.52 Exercise 2(e) I-2.54 Answers I-2.54 3. Application of Differential Calculus I-3.1 3.1 Curvature and Radius of Curvature I-3.1 3.2 Centre and Circle of Curvature I-3.6 Worked Example 3(a) I-3.7 Exercise 3(a) I-3.20 3.3 Evolutes and Envelopes I-3.21 Worked Example 3(b) I-3.23 3.4 Aliter I-3.33 Exercise 3(b) I-3.38 Answers I-3.40 4. Differential Calculus of Several Variables I-4.1 4.1 Introduction I-4.1 4.2 Total Differentiation I-4.1 Worked Example 4(a) I-4.4 Exercise 4(a) I-4.20 4.3 Jacobians I-4.27 4.4 Differentiation Under the Integral Sign I-4.31 Worked Example 4(b) I-4.33 Exercise 4(b) I-4.48 4.5 Maxima and Minima of Functions of Two Variables I-4.50 Worked Example 4(c) I-4.52 Exercise 4(c) I-4.65 Answers I-4.67 5. Multiple Integrals I-5.1 5.1 Introduction I-5.1 5.2 Evaluation of Double and Triple Integrals I-5.1 5.3 Region of Integration I-5.2 Worked Example 5(a) I-5.3 Exercise 5(a) I-5.15 5.4 Change of Order of Integration in a Double Integral I-5.17 5.5 Plane Area at Double Integral I-5.18 Worked Example 5(b) I-5.21 Exercise 5(b) I-5.38 5.6 Line Integral I-5.41 5.7 Surface Integral I-5.43 5.8 Volume Integral I-5.44 Contents vii Worked Example 5(c) I-5.45 Exercise 5(c) I-5.57 5.9 Gamma and Beta Functions I-5.59 Worked Example 5(d) I-5.63 Exercise 5(d) I-5.77 Answers I-5.79 Appendix A: Partial Differentiation I-A.1–21 Appendix B: Solutions to the January 2012 Question Paper I-B.1–9 Appendix C: Solutions to the January 2013 Question Paper I-C.1–13 Appendix D: Solutions to the Model Question Paper I I-D.1–9 Appendix E: Solutions to the Model Question Paper II I-E.1–9 Preface of my book on Engineering Mathematics-1 by the students and teachers throughout the country. course on Engineering Mathematics offered to the students of engineering. The contents have been covered in adequate depth for the semester 1 Syllabi of various universities/ deemed universities across the country. The book offers a balanced coverage of both theory and problems. Lucid writing style supported by step-by-step solutions to all problems enhances understanding of the concepts. It has an excellent pedagogy with 1212 unsolved problems, 539 solved problems, 669 short answer questions and 914 descriptive questions. For appendix. Additionally, the book is accompanied by a website which provides useful supplements including recent solved question papers. I hope that the book will be received by both the faculty and the students as enthusiastically as the earlier editions of the book and my other books. Critical evaluation and suggestions for the improvement of the book will be highly appreciated and acknowledged. T VEERARAJAN Publisher’s Note: McGraw Hill Education (India) Private Limited looks forward to receiving from teachers and students their valuable views, comments and suggestions for improvements, all of which may be sent to [email protected], mentioning the title and author’s name. PART I Mathematics-I 1. Matrices 2. Sequences and Series 3. Application of Differential Calculus 4. Differential Calculus of Several Variables 5. Multiple Integrals Chapter 1 Matrices 1.1 INTRODUCTION In the lower classes, the students have studied a few topics in Elementary Matrix theory. They are assumed to be familiar with the basic deﬁnitions and concepts of matrix theory as well as the elementary operations on and properties of matrices. Though the concept of rank of a matrix has been introduced in the lower classes, we brieﬂy recall the deﬁnition of rank and working procedure to ﬁnd the rank of a matrix, as it will be of frequent use in testing the consistency of a system of linear algebraic equations, that will be discussed in the next section. 1.1.1 Rank of a Matrix Determinant of any square submatrix of a given matrix A is called a minor of A. If the square submatrix is of order r, then the minor also is said to be of order r. Let A be an m × n matrix. The rank of A is said to be ‘r’, if (i) there is at least one minor of A of order r which does not vanish and (ii) every minor of A of order (r + 1) and higher order vanishes. In other words, the rank of a matrix is the largest of the orders of all the non- vanishing minors of that matrix. Rank of a matrix A is denoted by R(A) or ρ(A). To ﬁnd the rank of a matrix A, we may use the following procedure: We ﬁrst consider the highest order minor (or minors) of A. Let their order be r. If any one of them does not vanish, then ρ(A) = r. If all of them vanish, we next consider minors of A of next lower order (r – 1) and so on, until we get a non-zero minor. The order of that non-zero minor is ρ(A). This method involves a lot of computational work and hence requires more time, as we have to evaluate many determinants. An alternative method to ﬁnd the rank of a matrix A is given below: Reduce A to any one of the following forms, (called normal forms) by a series of elementary operations on A and then ﬁnd the order of the unit matrix contained in the normal form of A: Ir Ir | O Ir ; Ir | O ; ; O O|O I – 1.4 Part I: Mathematics I Here Ir denotes the unit matrix of order r and O is zero matrix. By an elementary operation on a matrix (denoted as E-operation) we mean any one of the following operations or transformations: (i) Interchange of any two rows (or columns). (ii) Multiplication of every element of a row (or column) by any non-zero scalar. (iii) Addition to the elements of any row (or column), the same scalar multiples of corresponding elements of any other row (or column). Note The alternative method for ﬁnding the rank of a matrix is based on the property that the rank of a matrix is unaltered by elementary operations. Finally we observe that we need not necessarily reduce a matrix A to the normal form to ﬁnd its rank. It is enough we reduce A to an equivalent matrix, whose rank can be easily found, by a sequence of elementary operations on A. The methods are illustrated in the worked examples that follow. 1.2 VECTORS A set of n numbers x1, x2,..., xn written in a particular order (or an ordered set of n numbers) is called an n-dimensional vector or a vector of order n. The n numbers are called the components or elements of the vector. A vector is denoted by a single letter X or Y etc. The components of a vector may be written in a row as X = (x1, x2,..., xn) x1 x2 or in a column as X. These are called respectively row vector and xn column vector. We note that a row vector of order n is a 1 × n matrix and a column vector of order n is an n × 1 matrix. 1.2.1 Addition of Vectors The sum of two vectors of the same dimension is obtained by adding the corresponding components. i.e., if X = (x1, x2,..., xn) and Y = (y1, y2,..., yn), then X + Y = (x1 + y1, x2 + y 2..., xn + yn). 1.2.2 Scalar Multiplication of a Vector If k is a scalar and X = (x1 , x2,..., xn) is a vector, then the scalar multiple kX is deﬁned as kX = (kx1, kx2,..., kxn). 1.2.3 Linear Combination of Vectors If a vector X can be expressed as X = k1X1 + k2X2 +... + krXr then X is said to be a linear combination of the vectors X1, X2,..., Xr. Chapter I: Matrices I – 1.5 1.3 LINEAR DEPENDENCE AND LINEAR INDEPENDENCE OF VECTORS The vectors X1, X2,..., Xr are said to be linearly dependent if we can ﬁnd scalars k1, k2,... kr, which are not all zero, such that k1Xl + k2X2 +... + krXr = 0. A set of vectors is said to be linearly independent if it is not linearly dependent, i.e. the vectors X1, X2,..., Xr are linearly independent, if the relation k1X1 + k2X2+... krXr = 0 is satisﬁed only when kl = k2 =... = kr = 0. Note When the vectors X1, X2,..., Xr are linearly dependent, then k1X1 + k2X2 +... + krXr = 0, where at least one of the k’s is not zero. Let km ≠ 0. k1 k2 kr Thus Xm X1 X2 Xr. km km km Thus at least one of the given vectors can be expressed as a linear combination of the others. 1.4 METHODS OF TESTING LINEAR DEPENDENCE OR INDEPENDENCE OF A SET OF VECTORS Method 1 Using the deﬁnition directly. Method 2 We write the given vectors as row vectors and form a matrix. Using elementary row operations on this matrix, we reduce it to echelon form, i.e. the one in which all the elements in the rth column below the rth element are zero each. If the number of non-zero row vectors in the echelon form equals the number of given vectors, then the vectors are linearly independent. Otherwise they are linearly dependent. Method 3 If there are n vectors, each of dimension n, then the matrix formed as in method (2) will be a square matrix of order n. If the rank of the matrix equals n, then the vectors are linearly independent. Otherwise they are linearly dependent. 1.5 CONSISTENCY OF A SYSTEM OF LINEAR ALGEBRAIC EQUATIONS Consider the following system of m linear algebraic equations in n unknowns: a11x1 + a12x2 +... + a1nxn = b1 a21x1 + a22x2 +... + a2nxn = b2 am1x1 + am2x2 +... + amnxn = bm This system can be represented in the matrix form as AX = B, where x1 b1 a11 a12 a1n x2 b2 A a21 a22 a2 n , X ,B am1 am 2 amn xn bm I – 1.6 Part I: Mathematics I The matrix A is called the coefﬁcient matrix of the system, X is the matrix of unknowns and B is the matrix of the constants. If B ≡ O, a zero matrix, the system is called a system of homogeneous linear equations; otherwise, the system is called a system of linear non-homogeneous equations. The m × (n + 1) matrix, obtained by appending the column vector B to the coefﬁcient matrix A as the additional last column, is called the augmented matrix of the system and is denoted by [A, B] or [A | B]. a11 a12 a1n b1 i.e. A, B a21 a22 a2 n b2 am1 am 2 amn bm 1.5.1 Deﬁnitions A set of values of x1, x2..., xn. which satisfy all the given m equations simultaneously is called a solution of the system. When the system of equations has a solution, it is said to be consistent. Otherwise the system is said to be inconsistent. A consistent system may have either only one or inﬁnitely many solutions. When the system has only one solution, it is called the unique solution. The necessary and sufﬁcient condition for the consistency of a system of linear non-homogeneous equations is provided by a theorem, called Rouches’s theorem, which we state below without proof. 1.5.2 Rouche’s Theorem The system of equations AX = B is consistent, if and only if the coefﬁcient matrix A and the augmented matrix [A, B] are of the same rank. Thus to discuss the consistency of the equations AX = B (m equations in n unknowns), the following procedure is adopted: We ﬁrst ﬁnd R(A) and R(A, B). (i) If R(A) ≠ R(A, B), the equations are inconsistent (ii) If R(A) = R(A, B) = the number of unknowns n, the equations are consistent and have a unique solution. In particular, if A is a non-singular (square) matrix, the system AX = B has a unique solution. (iii) If R(A) = R(A, B) < the number of unknowns n, the equations are consistent and have an inﬁnite number of solutions. 1.5.3 System of Homogeneous Linear Equations Consider the system of homogeneous linear equations AX = O (m equations in n unknowns) i.e a11x1 + a12x2 +... a1nxn = 0 a21x1 + a22x2 +... + a2nxn = 0 –––––––––––––––– am1x1 + am2x2 +... amnxn = 0 Chapter I: Matrices I – 1.7 This system is always consistent, as R(A) = R(A, O). If the coefﬁcient matrix A is non-singular, the system has a unique solution, namely, x1 = x2 =... = xn = 0. This unique solution is called the trivial solution, which is not of any importance. If the coefﬁcient matrix A is singular, i.e. if | A| = 0, the system has an inﬁnite number of non-zero or non-trivial solutions. The method of ﬁnding the non-zero solution of a system of homogeneous linear equations is illustrated in the worked examples that follow. WORKED EXAMPLE 1(a) Example 1.1 Show that the vectors X1 = (1, 1, 2), X2 = (1, 2, 5) and X3 = (5, 3, 4) are linearly dependent. Also express each vector as a linear combination of the other two. Method 1 Let k1X1 + k2X2 + k3X3 = 0 i.e. k1(1, 1, 2) + k2(1, 2, 5) + k3(5, 3, 4) = (0, 0, 0) ∴ k1 + k2 + 5k3 = 0 (1) k1 + 2k2 + 3k3 = 0 (2) 2k1 + 5k2 + 4k3 = 0 (3) (2) – (1) gives k2 – 2k3 = 0 or k2 =2k3 (4) Using (4) in (3), k1 = – 7k3 (5) Taking k3 = 1, we get k1 = – 7 and k2 = 2. Thus –7X1 + 2X2 + X3 = 0 (6) ∴ The vectors X1, X2, X3 are linearly dependent. 2 1 From (6), we get X1 X2 X3 , 7 7 7 1 X2 X1 X3 and X3 = 7X1 – 2X2 Method 2 2 2 Writing X1, X2, X3 as row vectors, we get 1 1 2 1 1 2 A 1 2 5 0 1 3 R2 R2 R1 , R3 R3 5 R1 5 3 4 0 2 6 1 1 2 0 1 3 R3 R3 2 R2 0 0 0 In the echelon form of the matrix, the number of non-zero vectors = 2 (< the number of given vectors). ∴ X1, X2, X3 are linearly dependent. I – 1.8 Part I: Mathematics I Now 0 R3 R3 2 R2 R3 5 R1 2 R2 R1 7 R1 2 R2 R3 i.e. –7X1 + 2X2 + X3 = 0 2 1 7 1 As before, X1 X2 X3 , X2 X1 X3 and X3 = 7X1 – 2X2. 7 7 2 2 Method 3 |A| = 0 ∴ R (A) ≠ 3; R (A) = 2 ∴ The vectors X1, X2, X3 are linearly dependent. Example 1.2 Show that the vectors X1 = (1, –1, –2, –4), X2 = (2, 3, –1, –1), X3 = (3, 1, 3, –2) and X4 = (6, 3, 0, –7) are linearly dependent. Find also the relationship among them. X1 1 1 2 4 1 1 2 4 X2 2 3 1 1 0 5 3 7 R2 R2 2 R1 , R3 A X3 3 1 3 2 0 4 9 10 R3 3R1 , R4 R4 6 R1 X4 6 3 0 7 0 9 12 17 1 1 2 4 3 7 0 1 1 5 5 R2 R2 ; R3 R3 ; R4 R4 5 0 4 9 10 0 9 12 17 1 1 2 4 3 7 0 1 5 5 33 22 R3 R3 4 R2 ; R4 R4 9 R2 0 0 5 5 33 22 0 0 5 5 1 1 2 4 3 7 0 1 5 5 R4 R4 R3 33 22 0 0 5 5 0 0 0 0 Number of non-zero vectors in echelon form of the matrix A = 3. ∴ The vectors X1, X2, X3, X4 are linearly dependent. Chapter I: Matrices I – 1.9 Now 0 R4 R4 R3 R4 9R2 R3 4R2 R4 R3 R2 R4 6R1 R3 3R1 R2 2R1 R1 R2 R3 R4 ∴ The relation among Xl, X2, X3, X4 is – X1 – X2 – X3 + X4 = 0 or X1 + X2 + X3 – X4 = 0. Example 1.3 Show that the vectors X1 = (2, –2, 1), X2 = (1, 4, –1) and X3 = (4, 6, –3) are linearly independent. Method 1 Let k1 X1 + k2 X2 + k3 X3 = 0 i.e. k1 (2, –2, 1) + k2 (1, 4, –1) + k3 (4, 6, –3) = (0, 0, 0) ∴ 2k1 + k2 + 4k3 = 0 (1) –2k1 + 4k2 + 6k3 = 0 (2) k1 – k2 –3k3 = 0 (3) From (1) and (2), k2 + 2k3 = 0 (4) From (2) and (3), k2 = 0 (5) ∴ k1 = 0 = k2 = k3. ∴ The vectors X1, X2, X3 are linearly independent. Method 2 X1 2 2 1 1 4 1 A X2 1 4 1 2 2 1 R1 R2 ; R2 R1 X3 4 6 3 4 6 3 1 4 1 0 10 3 R2 R2 2 R1 ; R3 R3 4 R1 0 10 1 1 1 4 0 3 R3 10 R3 R2 0 0 2 Number of non-zero vectors in the echelon form of A = number of given vectors, ∴ X1, X2, X3 are linearly independent. Example 1.4 Show that the vectors X1 = (1, −1, −1, 3), X2 = (2, 1, −2, −1) and X3 = (7, 2, −7, 4) are linearly independent. X1 1 1 1 3 1 1 1 3 R2 R2 2 R1 ; A X2 2 1 2 1 0 3 0 7 X3 7 2 7 4 0 9 0 17 R3 R3 7 R1 I – 1.10 Part I: Mathematics I 1 1 1 3 0 3 0 7 R3 R3 3R2 0 0 0 4 Number of non-zero vectors in the echelon form of A = number of given vectors. ∴ X1, X2, X3 are linearly independent. Example 1.5 Test for the consistency of the following system of equations: x1 + 2x2 + 3x3 + 4x4 = 5 6x1 + 7x2 + 8x3 + 9x4 = 10 11x1 + 12x2 + 13x3 + 14x4 =15 16x1 + 17x2+ 18x3 + 19x4 = 20 21x1 + 22x2 + 23x3 + 24x4 = 25 The given equations can be put as 1 2 3 4 5 x1 6 7 8 9 10 x2 11 12 13 14 15 x3 16 17 18 19 20 x4 21 22 23 24 25 i.e. AX = B (say) Let us ﬁnd the rank of the augmented matrix [A, B] by reducing it to the normal form 1 2 3 4 5 1 2 3 4 5 ( R2 R2 R1 6 7 8 9 10 5 5 5 5 5 R3 R3 R1 A, B 11 12 13 14 15 10 10 10 10 10 R4 R4 R1 16 17 18 19 20 15 15 15 15 15 R5 R5 R1 ) 21 22 23 24 25 20 20 20 20 20 Note If two matrices A and B are equivalent, i.e. are of the same rank, it is denoted as A ~ B. 1 2 3 4 5 1 1 1 1 1 1 1 1 R2 R2 , R3 R3 , R4 R4 , 5 10 15 1 1 1 1 1 1 1 1 1 1 1 R5 R5 20 1 1 1 1 1 1 2 3 4 5 0 1 2 3 4 R R2 R1 , R3 R3 R1 , 0 1 2 3 4 2 R R4 R1 , R5 R5 R1 0 1 2 3 4 4 0 1 2 3 4 Chapter I: Matrices I – 1.11 1 0 0 0 0 0 1 2 3 4 (C2 C2 2C1 , C3 C3 3C1 , 0 1 2 3 4 C4 C4 4C1 , C5 C5 5C1 ) 0 1 2 3 4 0 1 2 3 4 1 0 0 0 0 0 1 1 1 1 C2 C2 , C3 C3 2 , 0 1 1 1 1 C C4 3 , C5 C5 4 0 1 1 1 1 4 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 R R3 R2 , R4 R4 R2 , 0 0 0 0 0 3 R R5 R2 0 0 0 0 0 5 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 C C3 C2 , C4 C4 C2 , 0 0 0 0 0 3 C C5 C2 0 0 0 0 0 5 0 0 0 0 0 I2 0 Now [A, B] has been reduced to the normal form 0 0 The order of the unit matrix present in the normal form = 2. Hence the rank of [A, B] = 2. The rank of the coefﬁcient matrix A can be found as 2, in a similar manner. Thus R (A) = R [A, B] = 2 ∴ The given system of equations is consistent and possesses many solutions. Example 1.6 Test for the consistency of the following system of equations: x1 − 2x2 − 3x3 = 2; 3x1 − 2x2 = −1; −2x2 − 3x3 = 2; x2 + 2x3 = 1. The system can be put as 1 2 3 2 x1 3 2 0 1 x2 0 2 3 2 x3 0 1 2 1 I – 1.12 Part I: Mathematics I i.e. AX = B (say) 1 2 3 2 1 2 3 2 3 2 0 1 0 4 9 7 A, B R2 R2 3R1 0 2 3 2 0 2 3 2 0 1 2 1 0 1 2 1 1 0 0 0 0 4 9 7 C2 C2 2C1 , C3 C3 3C1 , C4 C4 2C1 0 2 3 2 0 1 2 1 1 0 0 0 0 1 2 1 R2 R4 , R4 R2 0 2 3 2 0 4 9 7 1 0 0 0 0 1 2 1 R3 R3 2 R2 , R4 R4 4 R2 0 0 1 4 0 0 1 11 1 0 0 0 0 1 0 0 C3 C3 2C2 , C4 C4 C1 0 0 1 4 0 0 1 11 1 0 0 0 0 1 0 0 R4 R4 R3 0 0 1 4 0 0 0 15 1 0 0 0 0 1 0 0 C4 C4 4C3 0 0 1 0 0 0 0 15 1 0 0 0 0 1 0 0 1 R4 R4 0 0 1 0 15 0 0 0 1 Chapter I: Matrices I – 1.13 ∴ R[A, B] = 4 But R (A) ≠ 4, as A is a (4 × 3) matrix. In fact R (A) = 3, as the value of the minor 3 2 0 0 2 3 0 0 1 2 Thus R (A) ≠ R [A, B] ∴The given system is inconsistent. Example 1.7 Test for the consistency of the following system of equations and solve them, if consistent, by matrix inversion. x − y + z + 1 = 0; x − 3y + 4z + 6 = 0; 4x + 3y − 2z + 3 = 0; 7x − 4y + 7z+ 16 = 0. 1 1 1 1 3 4 A 4 3 2 7 4 7 1 1 1 1 1 1 1 1 R2 R2 R1 , 1 3 4 6 0 2 3 5 A, B R3 R3 4 R1 , 4 3 2 3 0 7 6 1 R4 R4 7 R1 7 4 7 16 0 3 0 9 1 0 0 0 0 2 3 5 C2 C2 C1 , C3 C3 C1 , C4 C4 C1 0 7 6 1 0 3 0 9 1 0 0 0 0 1 6 7 R2 R3 and then C2 C4 0 5 3 2 0 9 0 3 1 0 0 0 R3 R3 5 R2 , R4 R4 9 R2 0 1 0 0 and then 0 0 27 33 C3 C3 6C2 , C4 C4 7C2 0 0 54 66 I – 1.14 Part I: Mathematics I 1 0 0 0 0 1 0 0 1 1 C3 C3 , C4 C4 0 0 1 1 27 33 0 0 2 2 1 0 0 0 0 1 0 0 R4 R4 2 R3 and theen 0 0 1 0 C4 C4 C3 0 0 0 0 ∴ R [A, B] = 3. Also R (A) = 3 ∴ The given system is consistent and has a unique solution. To solve the system, we take any three, say the ﬁrst three, of the given equations. 1 1 1 x 1 i.e. 1 3 4 y 6 4 3 2 z 3 i.e. AX = B, say ∴ X = A−1 B (1) 1 1 1 a11 a12 a13 Let A 1 3 4 a21 a22 a23 4 3 2 a31 a32 a33 Now A11 = co-factor of a11 in |A| = −6 A12 = 18; A13 = 15; A21 = 1; A22 = −6; A23 = −7; A31 = −1; A32 = −3; A33 = −2. 6 1 1 ∴ Adj A 18 6 3 15 7 2 |A| = a11 A11 + a12 A12 + a13 A13 = −9 6 1 1 ∴ 1 1 1 (2) A adj A 18 6 3 A 9 15 7 2 Using (2) in (1), x 6 1 1 1 1 y 18 6 3 6 9 z 15 7 2 3 Chapter I: Matrices I – 1.15 1 3 3 1 27 3 9 33 11 3 1 11 ∴ Solution of the system is x ,y 3, z 3 3 Example 1.8 Test for the consistency of the following system of equations and solve them, if consistent: 3x + y + z = 8; − x + y − 2z = − 5; x + y + z = 6; – 2x + 2y − 3z = − 7. Note As the solution can be found out by any method, when the system is consistent, we may prefer the triangularisation method (also known as Gaussian elimination method) to reduce the augmented matrix [A, B] to an equivalent matrix. Using the equivalent matrix, we can test the consistency of the system and also ﬁnd the solution easily when it exists. In this method, we use only elementary row operations and convert the elements below the principal diagonal of A as zeros. 3 1 1 8 1 1 1 6 1 1 2 5 1 1 2 5 A, B R1 R3 1 1 1 6 3 1 1 8 2 2 3 7 2 2 3 7 1 1 1 6 0 2 1 1 R2 R2 R1 , R3 R3 3R1 , R4 R4 2 R1 0 2 2 10 0 4 1 5 1 1 1 6 0 2 1 1 R3 R3 R2 , R4 R4 2 R2 0 0 3 9 0 0 1 3 1 1 1 6 0 2 1 1 1 R4 R4 R3 (1) 0 0 3 9 3 0 0 0 0 Now, Determinant of [A, B] = − Determinant of the equivalent matrix = 0. (∴ Two rows interchanged in the ﬁrst operation) ∴ R A, B 3 I – 1.16 Part I: Mathematics I 1 1 1 Now 0 2 1 6 0 0 0 3 ∴ R [A, B] = R (A) = 3 = the number of unknowns. ∴ The system is consistent and has a unique solution. A system of equations equivalent to the given system is also obtained from the equivalent matrix in (1). The equivalent equations are x + y + z = 6, 2y − z= 1 and − 3z = − 9 Solving them backwards, we get x = 1, y = 2, z = 3. Example 1.9 Examine if the following system of equations is consistent and ﬁnd the solution if it exists. x y z 1, 2 x 2y 3z 1; x y 2z 5; 3x y z 2. 1 1 1 1 1 1 1 1 R2 R2 2 R1 , 2 2 3 1 0 4 1 1 A, B R3 R3 R1 , 1 1 2 5 0 2 1 4 R4 R4 3R1 3 1 1 2 0 2 2 1 1 1 1 1 0 4 1 1 1 9 R 1 1 0 0 3 R3 R2 , R4 R4 R2 2 2 2 2 5 1 0 0 2 2 1 1 1 1 0 4 1 1 1 9 R4 R4 5 R3 0 0 2 2 0 0 0 22 It is obvious that det [A, B] = 4 and det [A] = 3 ∴ R [A, B] ≠ R [A]. ∴ The system is inconsistent. Note The last row of the equivalent matrix corresponds to the equation 0 x 0 y 0 z 22, which is absurd. From this also, we can conclude that the system is inconsistent. Chapter I: Matrices I – 1.17 Example 1.10 Solve the following system of equations, if consistent: x + y + z = 3, x + y − z = 1; 3x + 3y − 5z = 1. 1 1 1 3 1 1 1 3 A, B 1 1 1 1 0 0 2 2 R2 R2 R1 , R3 R3 3R1 3 3 5 1 0 0 8 8 1 1 1 3 0 0 2 2 R3 R3 4 R2 0 0 0 0 ∴ All the third order determinants vanish ∴ R [A, B] ≠ 3 1 1 Consider , which is a minor of both A and [A, B]. 0 2 The value of this minor = − 2 ≠ 0 ∴ R (A) = R [A, B] < the number of unknowns. ∴ The system is consistent with many solutions. From the ﬁrst two rows of the equivalent matrix, we have x + y + z = 3 and − 2z = − 2 i.e. z=1 and x + y = 2. ∴ The system has a one parameter family of solutions, namely x = k, y = 2 − k, z = 1, where k is the parameter. Giving various values for k, we get inﬁnitely many solutions. Example 1.11 Solve the following system of equations, if consistent: x1 2 x2 x3 5 x4 4; x1 3 x2 2 x3 7 x4 5; 2 x1 x2 3 x3 3. 1 2 1 5 4 A, B 1 3 2 7 5 2 1 3 0 3 1 2 1 5 4 0 1 1 2 1 R2 R2 R1 , R3 R3 2 R1 0 5 5 10 0 5 1 2 1 5 4 0 1 1 2 1 R3 R3 5 R2 0 0 0 0 0 ∴ ∴ R [A, B] ≠ 3 ( the last row contains only zeros) Similarly R (A) ≠ 3. 1 2 Since 0 , R (A) = R [A, B] = 2 < the number of unknowns. 0 1 ∴ The given system is consistent with many solutions. I – 1.18 Part I: Mathematics I From the ﬁrst two rows of the equivalent matrix, we have x1 2 x2 x3 5 x4 4 (1) and x2 x3 2 x4 1 (2) As there are only 2 equations, we can solve for only 2 unknowns. Hence the other 2 unknowns are to be treated as parameters. Taking x3 = k and x4 = k′, we get x2 = 1 + k + 2k' [from (2)] and x1 = 4 − 2 (1+ k + 2k') + k + 5k' [from (1)] i.e. x1 = 2 − k + k' ∴ The given system possesses a two parameter family of solutions. Note From the Examples (10) and (11), we note that the number of parameters in the solution equals the difference between the number of unknowns and the common rank of A and [A, B]. Example 1.12 Find the values of k, for which the equations x + y + z = 1, x + 2y + 3z = k and x + 5y + 9z = k2 have a solution. For these values of k, ﬁnd the solutions also. 1 1 1 1 1 1 1 1 R R2 R1 , A, B 1 2 3 k 0 1 2 k 1 2 R R3 R1 1 5 9 k2 0 4 8 k2 1 3 1 1 1 1 0 1 2 k 1 R3 R3 4 R2 (1) 2 0 0 0 k 4k 3 1 1 1 A 0 1 2 R A 2 0 0 0 If the system possesses a solution, R [A, B] must also be 2. ∴ The last row of the matrix in (1) must contain only zeros. ∴ k2 − 4k + 3 = 0 i.e. k = 1 or 3. For these values of k, R (A) = R [A, B] = 2 < the number of unknowns. ∴ The given system has many solutions. Case (i) k=1 The ﬁrst two rows of (1) give the equivalent equations as and x+y+z=1 (2) y + 2z = 0 (3) Puting z = λ, the one-parameter family of solutions of the given system is x = λ + 1, y = − 2λ and z=λ Case (ii) k=3 The equivalent equations are x+y+z=1 (2) Chapter I: Matrices I – 1.19 and y + 2z = 2 (4) Putting z = μ, the one-parameter family of solutions of the given system is x = μ − 1, y = 2 − 2 μ, z = μ. Example 1.13 Find the condition satisﬁed by a, b, c, so that the following system of equations may have a solution: x + 2y − 3z = a; 3x − y + 2z = b; x − 5y + 8z = c. 1 2 3 a A, B 3 1 2 b 1 5 8 c 1 2 3 a 0 7 11 b 3a R2 R2 3R1 , R3 R3 R1 0 7 11 c a 1 2 3 a 0 7 11 b 3a R3 R3 R2 (1) 0 0 0 2a b c 1 2 3 A 0 7 11 R A 2 0 0 0 If the given system possesses a solution, R [A, B] = 2. ∴ The last row of (1) should contain only zeros. ∴ 2a − b + c = 0. Only when this condition is satisﬁed by a, b, c, the system will have a solution. Example 1.14 Find the value of k such that the following system of equations has (i) a unique solution, (ii) many solutions and (iii) no solution. kx + y + z = 1; x + ky + z = 1; x + y + kz = 1. k 1 1 A 1 k 1 ∴ 1 1 k A k k2 1 1 k 1 k 2 k 1 k k 2 2 k 1 k 2 |A| = 0, when k = 1 or k = –2 When k ≠ 1 and k ≠ −2, |A| ≠ 0 ∴ R(A) = 3 Then the system will have a unique solution. I – 1.20 Part I: Mathematics I When k = 1, the system reduces to the single equation x + y + z = 1. In this case, R(A) = R[A, B] = 1. ∴ The system will have many solutions. (i.e. a two parameter family of solutions) When k = − 2, 2 1 1 1 1 2 1 1 A, B 1 2 1 1 2 1 1 1 R1 R2 1 1 2 1 1 1 2 1 1 2 1 1 0 3 3 3 R2 R2 2 R1 , R3 R3 R1 0 3 3 0 1 2 1 1 0 3 3 3 R3 R3 R2 0 0 0 3 Now 1 2 1 0 3 3 0 R A 3 0 0 0 1 2 0 R A 2 0 3 2 1 1 3 3 3 a minor of A, B 0 0 0 3 ∴ R[A, B] = 3. Thus R(A) ≠ R[A, B]. ∴ The system has no solution. Example 1.15 Investigate for what values of λ, μ, the equations x + y + z = 6, x + 2y + 3z = 10 and x + 2y + λz = μ have (i) no solution, (ii) a unique solution, (iii) an inﬁnite number of solutions. 1 1 1 6 1 1 1 6 R2 R2 R1 , A, B 1 2 3 10 0 1 2 4 R3 R3 R1 1 2 0 1 1 6 1 1 1 6 0 1 2 4 R3 R3 R2 0 0 3 10 1 1 1 A 0 1 2 and A 3 0 0 3 Chapter I: Matrices I – 1.21 If λ ≠ 3, |A| ≠ 0 ∴ R(A) = 3 ∴ When λ ≠ 3 and μ takes any value, the system has a unique solution. 1 1 If λ = 3, |A| = 0 and a second order minor of A, i.e. 0 0 1 ∴ R (A) = 2. 1 1 1 6 When λ = 3, A, B 0 1 2 4 (1) 0 0 0 10 When λ = 3 and μ = 10, the last row of (1) contains only zeros. ∴ R[A, B] ≠ 3 and clearly R[A, B] = 2. Thus, when λ = 3 and μ = 10, R(A) = R[A, B] = 2. ∴ The system has an inﬁnite number of solutions. When λ = 3 and μ ≠ 10, a third order minor of [A, B], i.e. 1 1 6 1 2 4 10 0 0 0 10 ∴ R [A, B] = 3 Thus, when λ = 3 and μ ≠ 10, R(A) ≠ R[A, B]. ∴ The given system has no solution. Example 1.16 Test whether the following system of equations possess a non-trivial solution. x1 + x2 + 2x3 + 3x4 = 0; 3x1 + 4x2 + 7x3 + 10x4 = 0; 5x1 + 7x2 + 11x3 + 17x4 = 0; 6x1 + 8x2 + 13x3 + 16x4 = 0. The given system is a homogeneous linear system of the form AX = 0. 1 1 2 3 1 1 2 3 R2 R2 3R1 , 3 4 7 10 0 1 1 1 A R3 R3 5 R1 , 5 7 11 17 0 2 1 2 R4 R4 6 R1 6 8 13 16 0 2 1 2 1 1 2 3 0 1 1 1 R3 R3 2 R2 , R4 R4 2 R2 0 0 1 0 0 0 1 4 1 1 2 3 0 1 1 1 R4 R4 R3 0 0 1 0 0 0 0 4 ∴ |A| = 4 i.e. A is non-singular I – 1.22 Part I: Mathematics I R(A) = R[A, 0] = 4 ∴ The system has a unique solution, namely, the trivial solution. Example 1.17 Find the non-trivial solution of the equations x + 2y + 3z = 0, 3x + 4y + 4z = 0, 7x + 10y + 11z = 0, if it exists. 1 2 3 1 2 3 R R2 3R1 , A 3 4 4 0 2 5 2 R3 R3 7 R1 7 10 11 0 4 10 1 2 3 0 2 5 R3 R3 2 R2 (1) 0 0 0 1 2 ∴ |A| = 0 and 0 ∴ R(A) = 2 0 2 ∴ The system has non-trivial solution. From the ﬁrst two rows of (1), we see that the given equations are equivalent to x + 2y + 3z = 0 (2) and − 2y − 5z = 0 (3) 5 Putting z = k, we get y k from (3) and x = 2k. 2 Thus the non-trivial solution is x = 4k, y = –5k and z = 2k. Example 1.18 Find the non-trivial solution of the equations x − y + 2z − 3w = 0, 3x + 2y – 4z + w = 0, 5x – 3y + 2z + 6w = 0, x – 9y + 14z − 2w = 0, if it exists. 1 1 2 3 1 1 2 3 R2 R2 3R1 , 3 2 4 1 0 5 10 10 A R3 R3 5 R1 , 5 3 2 6 0 2 8 21 R4 R4 R1 1 9 14 2 0 8 12 1 1 1 2 3 0 1 2 2 1 R2 R2 0 2 8 21 5 0 8 12 1 1 1 2 3 0 1 2 2 R3 R3 2 R2 , R4 R4 8 R2 0 0 4 17 0 0 4 17 Chapter I: Matrices I – 1.23 1 1 2 3 0 1 2 2 ~ R4 R4 R3 0 0 4 17 0 0 0 0 ∴ |A| = 0 i.e. R(A) < 4 ∴ The system has a non-trivial solution. The system is equivalent to x − y + 2z − 3w = 0 (1) y − 2z +2w = 0 (2) −4z + 17w = 0 (3) Putting w = 4k, we get z = 17k from (3), y = 26k from (2) and x = 4k. Thus the non-trivial solution is x = 4k, y = 26k, z = 17k and w = 4k. Example 1.19 Find the values of λ for which the equations x +(λ + 4) y + (4λ + 2)z = 0, x + 2(λ + 1) y + (3λ + 4) z = 0, 2x + 3λy + (3λ + 4) z = 0 have a non-trivial solution. Also ﬁnd the solution in each case. 1 4 4 2 A 1 2 2 3 4 2 3 3 4 1 4 4 2 R2 R2 R1 , ~ 0 2 2 R3 R3 2 R1 (1) 0 8 5 For non-trivial solution, |A| = 0 i.e. −5λ (λ − 2) − (λ − 8) (2 − λ) = 0 i.e. −4λ2 + 16 = 0 ∴ λ=±2 When λ = 2, the system is equivalent to x + 6y + 10z = 0 −6y − 10z = 0, from (1) Putting z = 3k, we get y = −5k and x = 0 i.e. the solution is x = 0, y = −5k and z = 3k. When λ = −2, the system is equivalent to x + 2y − 6z = 0 −4y + 4z = 0, from (1) Putting z = k, we get y = k and x = 4k. i.e. the solution is x = 4k, y = k and z = k. I – 1.24 Part I: Mathematics I EXERCISE 1(a) Part A (Short Answer Questions) 1. Deﬁne the linear dependence of a set of vectors. 2. Deﬁne the linear independence of a set of vectors. 3. If a set of vectors is linearly dependent, show that at least one member of the set can be expressed as a linear combination of the other members. 4. Show that the vectors X1 = (1, 2), X2 = (2, 3) and X3 = (4, 5) are linearly dependent. 5. Show that the vectors X1 = (0, 1, 2), X2 = (0, 3, 5) and X3 = (0, 2, 5) are linearly dependent. 6. Express X1 = (1, 2) as a linear combination of X2 = (2, 3) and X3 = (4, 5). 7. Show that the vectors (1, 1, 1), (1, 2, 3) and (2, 3, 8) are linearly independent. 8. Find the value of a if the vectors (2, −1, 0), (4, 1, 1) and (a, −1, 1) are linearly dependent. 9. What do you mean by consistent and inconsistent systems of equations. Give examples. 10. State Rouche’s theorem. 11. State the condition for a system of equations in n unknowns to have (i) one solution, (ii) many solutions and (iii) no solution. 12. Give an example of 2 equations in 2 unknowns that are (i) consistent with only one solution and (ii) inconsistent. 13. Give an example of 2 equations in 2 unknowns that are consistent with many solutions. 14. Find the values of a and b, if the equations 2x − 3y = 5 and ax + by = −10 have many solutions. 15. Test if the equations x + y + z = a, 2x + y + 3z = b, 5x + 2y + z = c have a unique solution, where a, b, c are not all zero. 16. Find the value of λ, if the equations x + y − z = 10, x − y + 2z = 20 and λx − y + 4z = 30 have a unique solution. 17. If the augmented matrix of a system of equations is equivalent to 1 2 1 2 0 5 3 2 , ﬁnd the value of λ, for which the system has a unique 0 0 0 solution. 18. If the augmented matrix of a system of equations is equivalent to 1 2 1 3 0 2 2 2 , ﬁnd the values of λ and μ for which the system has 0 0 1 3 only one solution. Chapter I: Matrices I – 1.25 19. If the augmented matrix of a system of equations is equivalent to 1 2 3 4 0 5 4 2 , ﬁnd the values of λ and μ for which the system 0 0 2 3 has many solutions. 20. If the augmented matrix of a system of equations is equivalent to 1 1 2 3 0 3 1 2 , ﬁnd the values of λ and μ for which the system 0 0 8 11 has no solution. 21. Do the equations x − 3y − 8z = 0, 3x + y = 0 and 2x + 5y + 6z = 0 have a non- trivial solution? Why? 22. If the equations x + 2y + z = 0, 5x + y –z = 0 and x + 5y + λz = 0 have a non- trivial solution, ﬁnd the value of λ. 23. Given that the equations x + 2y − z = 0, 3x + y − z = 0 and 2x – y = 0 have non-trivial solution, ﬁnd it. Part B Show that the following sets of vectors are linearly dependent. Find their relationship in each case: 24. X1 = (1, 2, 1), X2 = (4, 1, 2), X3 = (6, 5, 4), X4 = (−3, 8, 1). 25. X1 = (3, 1, −4), X2 = (2, 2, −3), X3 = (0, −4, 1), X4 = (−4, −4, 6) 26. X1 = (1, 2, −1, 3), X2 = (0, −2, 1, −1), X3 = (2, 2, −1, 5) 27. X1 = (1, 0, 4, 3), X2 = (2, 1, −1, 1), X3 = (3, 2, −6, −1) 28. X1 = (1, −2, 4, 1), X2 = (1, 0, 6, −5), X3 = (2, −3, 9, −1) and X4 = (2, −5, 7, 5). 29. Determine whether the vector x5 = (4, 2, 1, 0) is a linear combination of the set of vectors X1 = (6, − 1, 2, 1), X2 = (1, 7, − 3, −2), X3 = (3, 1, 0, 0) and X4 = (3, 3,−2,−1). Show that each of the following sets of vectors is linearly independent. 30. X1 = (1, 1, 1); X2 = (1, 2, 3); X3 = (2, −1, 1). 31. X1 = (1, −1, 2, 3); X2 = (1, 0, −1, 2); X3 = (1, 1, −4, 0) 32. X1 = (1, 2, −1, 0) X2 = (1, 3, 1, 2); X3 = (4, 2, 1, 0); X4 = (6, 1, 0, 1). 33. X1 = (1, −2, −3, −2, 1); X2 = (3, −2, 0, −1, −7); X3 = (0, 1, 2, 1, −6); X4 = (0, 2, 2, 1, −5). 34. Test for the consistency of the following system of equations: 3 4 5 6 7 x1 4 5 6 7 8 x2 5 6 7 8 9 x3 10 11 12 13 14 x4 15 16 17 18 19 I – 1.26 Part I: Mathematics I Test for the consistency of the following systems of equations and solve, if con- sistent: 35. 2x − 5y + 2z = −3; −x − 3y + 3z = −1; x + y − z = 0; −x + y = 1. 36. 3x + 5y − 2z = 1; x − y + 4z = 7; −6x − 2y + 5z = 9; 7x − 3y + z = 4. 37. 2x + 2y + 4z = 6; 3x + 3y + 7z = 10; 5x + 7y + 11z = 17; 6x + 8y + 13z = 16. Test for the consistency of the following systems of equations and solve, if consistent: 38. x + 2y + z = 3; 2x + 3y + 2z = 5; 3x − 5y + 5z = 2; 3x + 9y − z = 4. 39. 2x + 6y − 3z = 18; 3x − 4y + 7z = 31; 5x + 3y + 3z = 48; 8x − 3y + 2z = 21. 40. x + 2y + 3z = 6; 5x − 3y + 2z = 4; 2x + 4y − z = 5; 3x + 2y + 4z = 9. 41. x + 2y = 4; 10y + 3z = −2; 2x − 3y − z = 5; 3x + 3y + 2z = 1. 42. 2x1 + x2 + 2x3 + x4 = 6; x1 − x2 + x3 + 2x4 = 6; 4x1 + 3x2 + 3x3 − 3x4 = −1; 2x1 + 2x2 − x3 + x4= 10 43. 2x + y + 5z + w = 5; x + y + 3z − 4w = − 1; 3x + 6y − 2z + w = 8; 2x + 2y + 2z − 3w = 2. Test for the consistency of the following systems of equations and solve, if consistent: 44. x − 3y − 8z = − 10; 3x + y = 4z; 2x + 5y + 6z = 13. 45. 5x + 3y + 7z = 4; 3x + 26y + 2z = 9; 7x + 2y + 10z = 5. 46. x − 4y − 3z + 16 = 0; 2x + 7y + 12z = 48; 4x − y + 6z = 16; 5x − 5y + 3z = 0. 47. x − 2y + 3w = 1; 2x − 3y + 2z + 5w = 3; 3x − 7y − 2z + 10w = 2. 48. x1 + 2x2 + 2x3 − x4 = 3; x1 + 2x2 + 3x3 + x4 = 1; 3x1 + 6x2 + 8x3 + x4 = 5. 49. Find the values of k, for which the equations x + y + z = 1, x + 2y + 4z = k and x + 4y + 10z = k2 have a solution. For these values of k, ﬁnd the solutions also. 50. Find the values of λ, for which the equtions x + 2y + z = 4, 2x − y − z = 3λ and 4x − 7y − 5z = λ2 have a solution. For these values of λ, ﬁnd the solutions also. 51. Find the condition on a, b, c, so that the equations x + y + z = a, x + 2y + 3z = b, 3x + 5y + 7z = c may have a one-parameter family of solutions. 52. Find the value of k for which the equations kx − 2y + z = 1, x − 2ky + z = −2 and x − 2y + kz = 1 have (i) no solution, (ii) one solution and (iii) many solutions. 53. Investigate for what values of λ, μ the equations x + y + 2z = 2, 2x − y + 3z = 2 and 5x − y + λz = μ have (i) no solution, (ii) a unique solution, (iii) an inﬁnite number of solutions. 54. Find the values of a and b for which the equations x + y + 2z = 3, 2x − y + 3z = 4 and 5x − y + az = b have (i) no solution, (ii) a unique solution, (iii) many solutions. 55. Find the non-trivial solution of the equations x + 2y + z = 0; 5x + y − z = 0 and x + 5y + 3z = 0, if it exists. 56. Find the non-trivial solution of the equations x + 2y + z + 2w = 0; x + 3y + 2z + 2w = 0; 2x + 4y + 3z + 6w = 0 and 3x + 7y + 4z + 6w = 0, if it exists. 57. Find the values of λ for which the equations 3x + y − λz = 0, 4x − 2y − 3z = 0 and 2λx + 4y + λz = 0 possess a non-trivial solution. For these values of λ, ﬁnd the solution also. 58. Find the values of λ for which the equations (11 − λ) x − 4y − 7z = 0, 7x − (λ + 2) y − 5z = 0, 10x − 4y − (6 + λ) z = 0 possess a non-trivial solution. For these values of λ, ﬁnd the solution also. Chapter I: Matrices I – 1.27 1.6 EIGENVALUES AND EIGENVECTORS 1.6.1 Deﬁnition Let A = [aij] be a square matrix of order n. If there exists a non-zero column vector X and a scalar λ, such that AX = λX then λ is called an eigenvalue of the matrix A and X is called the eigenvector corresponding to the eigenvalue λ. To ﬁnd the eigenvalues and the corresponding eigenvectors of a square matrix A, we proceed as follows: Let λ be an eigenvalue of A and X be the corresponding eigenvector. Then, by deﬁnition, AX = λX = λIX, where I is the unit matrix of order n. i.e. (A − λI) X = 0 (1) x1 0 a11 a12 a1n 1 0 0 x2 0 i.e. a21 a22 a2 n 0 1 0 an1 an 2 ann 0 0 0 1 xn 0 i.e. a11 x1 a12 x2 a1n xn 0 a21 x1 a22 x2 a2 n xn 0 (2)..................................................... an1 x1 an 2 x2 ann xn 0 Equations (2) are a system of homogeneous linear equations in the unknowns x1, x2,... , xn. x1 x2 Since X is to be a non-zero vector, xn x1, x2,... , xn should not be all zeros. In other words, the solution of the system (2) should be a non-trivial solution. The condition for the system (2) to have a non-trivial solution is a11 a12 a1n a21 a22 a2 n 0 (3) an1 an 2 ann i.e. |A − λ I| = 0 (4) The determinant |A − λI| is a polynomial of degree n in λ and is called the characteristic polynomial of A. The equation |A − λ I| = 0 or the equation (3) is called the characteristic equation of A. I – 1.28 Part I: Mathematics I When we solve the characteristic equation, we get n values for λ. These n roots of the characteristic equation are called the characteristic roots or latent roots or eigenvalues of A. Corresponding to each value of λ, the equations (2) possess a non-zero (non- trivial) solution X. X is called the invariant vector or latent vector or eigenvector of A corresponding to the eigenvalue λ. Notes 1. Corresponding to an eigenvalue, the non-trivial solution of the system (2) will be a one-parameter family of solutions. Hence the eigenvector corresponding to an eigenvalue is not unique. 2. If all the eigenvalues λ1, λ2,... , λn of a matrix A are distinct, then the cor- responding eigenvectors are linearly independent. 3. If two or more eigenvalues are equal, then the eigenvectors may be linearly independent or linearly dependent. 1.6.2 Properties of Eigenvalues 1. A square matrix A and its transpose AT have the same eigenvalues. Let A = (aij); i, j = 1, 2,... , n. The characteristic polynomial of A is a11 a12 a1n A I a21 a22 a2 n (1) an1 an 2 ann T The characteristic polynomial of A is a11 a21 an1 T A I a12 a22 an 2 (2) a1n a2 n ann Determinant (2) can be obtained by changing rows into columns of determi- nant (1). ∴ |A − λ I| = |AT − λ I| ∴ The characteristic equations of A and AT are identical. ∴ The eigenvalues of A and AT are the same. 2. The sum of the eigenvalues of a matrix A is equal to the sum of the principal diagonal elements of A. (The sum of the principal diagonal elements is called the Trace of the matrix.) The characteristic equation of an nth order matrix A may be written as n n 1 n 2 n D1 D2 1 Dn 0, (1) where Dr is the sum of all the rth order minors of A whose principal diagonals lie along the principal diagonal of A. Chapter I: Matrices I – 1.29 (Note Dn = |A|). We shall verify the above result for a third order matrix. a11 a12 a13 Let A a21 a22 a23 a31 a32 a33 The characteristic equation of A is given by a11 a12 a13 a21 a22 a23 0 (2) a31 a32 a33 Expanding (2), the characteristic equation is 2 a22 a23 a11 a22 a33 a32 a33 a21 a23 a21 a22 a12 a21 a13 a31 0 a31 a33 a31 a32 3 2 i.e. a11 a22 a33 a11 a12 a11 a13 a22 a23 A 0 a21 a22 a31 a33 a32 a33 i.e. λ3 − D1 λ2 + D2 λ − D3 = 0, using the notation given above. This result holds good for a matrix of order n. Note This form of the characteristic equation provides an alternative method for getting the characteristic equation of a matrix. Let λ1, λ2,... , λn be the eigenvalues of A. ∴ They are the roots of equation (1). D1 ∴ 1 2 n D1 1 a11 a22 ann Trace of the matrix A. 3. The product of the eigenvalues of a matrix A is equal to |A|. If λ1, λ2,... , λn are the eigenvalues of A, they are the roots of n n 1 n 2 n D1 D2 1 Dn 0. n n 1 1 Dn ∴ Product of the roots 1 i.e. λ1, λ2... λn = Dn = |A|. 1.6.3 Aliter λ1, λ2,... , λn are the roots of |A − λI| = 0 I – 1.30 Part I: Mathematics I ∴ |A − λI| ≡ (−1)n (λ − λ1) (λ − λ2)... (λ − λn), since L.S. is a nth degree polynomial in λ whose leading term is (−1)n λn. Putting λ = 0 in the above identity, we get |A| = (−1)n (−λ1) (−λ2)... (−λn) i.e. λ1 λ2... λn = |A|. 1.6.4 Corollary If |A| = 0, i.e. A is a singular matrix, at least one of the eigenvalues of A is zero and conversely. 4. If λ1, λ2,..., λn are the eigenvalues of a matrix A, then (i) kλ1, kλ2,... kλn are the eigenvalues of the matrix kA, where k is a non- zero scalar. (ii) 1p , 2p , , np are the eigenvalues of the matrix Ap, where p is a positive integer. (iii) 1 , 1 , 1 are the eigenvalues of the inverse matrix A−1, provided λr 1 2 n ≠ 0 i.e. A is non-singular. (i) Let λr be an eigenvalue of A and Xr the corresponding eigenvector. Then, by deﬁnition, AXr = λr xr (1) Multiplying both sides of (1) by k, (kA)Xr = (kλr) Xr (2) From (2), we see that kλr is an eigenvalue of kA and the corresponding eigenvector is the same as that of λr, namely Xr. (ii) Premultiplying both sides of (1) by A, A2 X r A AX r A r Xr r AX r 2 r Xr 3 3 Similarly A X r r X r and so on. In general, A p X r p r Xr p From, (3), we see that r is an eigenvalue of Ap with the corresponding eigenvector equal to Xr, which is the same for λr. (iii) Premultiplying both sides of (1) by A−1, A−1 (AXr) = A−1 (λr Xr) i.e. Xr = λr (A−1 Xr) 1 ∴ A 1Xr Xr (4) r Chapter I: Matrices I – 1.31 1 From (4), we see that is an eigenvalue of A–1 with the corresponding r eigenvector equal to Xr, which is the same for λr. 5. The eigenvalues of a real symmetric matrix (i.e. a symmetric matrix with real elements) are real. Let λ be an eigenvalue of the real symmetric matrix and X be the corre- sponding eigenvector. Then AX = λ (1) Premultiplying both sides of (1) by X T (the transpose of the conjugate of X), we get X T AX XT X (2) Taking the complex conjugate on both sides of (2), XT A X X T X (assuming that may be complex ) i.e. XT AX XT X ( A A, as A is real) (3) Taking transpose on both sides of (3), X T AT X XT X ( AB )T BT AT i.e. XT AX XT X ( A)T A, as A issymmetric (4) From (2) and (4), we get XT X XT X ( )XT X 0 i.e. X T X is an 1 × 1 matrix, i.e. a single element which is positive ∴ 0 i.e. λ is real. Hence all the eigenvalues are real. 6. The eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are orthogonal. x1 y1 x2 y2 Note Two column vectors X and Y are said to be xn yn orthogonal, if their inner product (x1y1 + x2y2 + … xnyn) = 0 i.e. if XTY = 0. Let λ1, λ2 be any two distinct eigenvalues of the real symmetric matrix A and X1, X2 be the corresponding eigenvectors respectively. Then AX1 = λ1X1 (1) and AX2 = λ2X2 (2) Premultiplying both sides of (1) by, X 2T we get X 2T AX 1 1 X 2T X 1 I – 1.32 Part I: Mathematics I Taking the transpose on both sides, X 1T AX 2 1 X 1T X 2 ( AT A) (3) Premultiplying both sides of (2) by X1T , we get X 1T AX 2 2 X 1T X 2 (4) From (3) and (4), we have 1 X 1T X 2 2 X 1T X 2 T ( 1 2 X1 X 2 ) 0 i.e. 1 2 , X 1T X 2 0 Since i.e. the eigenvectors X1 and X2 are orthogonal. WORKED EXAMPLE 1(b) 5 4 Example 1.1 Given that A , verify that the eigenvalues of A2 are the 1 2 squares of those of A. Verify also that the respective eigenvectors are the same. 5 4 The characteristic equation of A is 0 1 2 i.e. (5 − λ) (2 − λ) − 4 = 0 i.e. λ2 − 7λ + 6 = 0 ∴ The eigenvalues of A are λ = 1, 6. The eigenvector corresponding to any λ is given by (A − λ I) X = 0 5 4 x1 i.e. 0 1 2 x2 When λ = 1, the eigenvector is given by the equations 4x1 + 4x2 = 0 and x1 + x2 = 0, which are one and the same. Solving, x1 = − x2. Taking x1 = 1, x2 = −1. ∴ The eigenvector is 1 1 Chapter I: Matrices I – 1.33 When λ = 6, the eigenvector is given by −x1 + 4x2 = 0 and x1 − 4x2 = 0 Solving, x1 = 4x2 Taking x2 = 1, x1 = 4 1 ∴ The eigenvector is 4 5 4 5 4 29 28 Now A2 1 2 1 2 7 8 29 28 The characteristic equation of A2 is 0 7 8 i.e. (29 − λ) (8 − λ) − 196 = 0 i.e. λ2 − 37λ + 36 = 0 i.e. (λ − 1)(λ − 36) = 0 ∴ The eigenvalues of A2 are 1 and 36, that are the squares of the eigenvalues of A, namely 1 and 6. When λ = 1, the eigenvector of A2 is given by 28 28 x1 0. i.e. 28x1 28 x2 0 and 7 x1 7 x2 0 7 7 x2 Solving, x1 = –x2. Taking x1 = 1, x2 = –1. When λ = 36, the eigenvector of A2 is given by 7 28 x1 0. i.e. 7 x1 28 x2 0 and 7 x1 28 x2 0. 7 28 x2 Solving, x1 = 4x2. Taking x2 = 1, x1 = 4. Thus the eigenvectors of A2 are 1 4 and , which are the same as the respective eigenvectors of A. 1 1 Example 1.2 Find the eigenvalues and eigenvectors of the matrix 1 1 3 A 1 5 1 3 1 1 The characteristic equation of A is 1 1 3 1 5 1 0 3 1 1 i.e. (l – λ) {λ2 – 6λ + 4} – (1 – λ – 3) + 3(1 – 15 + 3λ) = 0 i.e. –λ3 + 7λ2 – 36 = 0 or λ3 – 7λ2 + 36 = 0 (1) i.e. (λ + 2) (λ2 – 9λ + 18) = 0 [∴ λ= –2 satisﬁes (1) ] i.e. (λ + 2) (λ – 3) (λ – 6) = 0 ∴ The eigenvalues of A are λ = –2, 3, 6. Case (i) λ = –2. The eigenvector is given by I – 1.34 Part I: Mathematics I 3 1 3 x1 1 7 1 x2 0 (2) 3 1 3 x3 i.e. x1 + 7x2 + x3 = 0 3x1 + x2 + 3x3 = 0 Solving these equations by the rule of cross-multiplication, we have x1 x2 x3 21 1 3 3 1 21 x1 x2 x3 (3) 20 0 20 i.e. Note To solve for x1, x2, x3, we have taken the equations corresponding to the second and third rows of the matrix in step (2). The proportional values of x1, x2, x3 obtained in step (3) are the co-factors of the elements of the ﬁrst row of the determinant of the matrix in step (2). This provides an alternative method for ﬁnding the eigenvector. From step (3), x1 = k, x2 = 0 and x3 = –k. Usually the eigenvector is expressed in terms of the simplest possible numbers, corresponding to k = 1 or – 1. ∴ x1 = 1, x2 = 0, x3 = – 1 Thus the eigenvector corresponding to λ = – 2 is 1 X1 0 1 Case (ii) λ = 3. 2 1 3 x1 The eigenvector is given by 1 2 1 x2 0. 3 1 2 x3 Values of x1, x2, x3 are proportional to the co-factors of –2, 1, 3 (elements of the ﬁrst row i.e. –5, 5, –5. x1 x2 x3 x1 x2 x3 i.e. or 5 5 5 1 1 1 1 ∴ X2 1 1 Case (iii) λ = 6. 5 1 3 x1 The eigenvector is given by 1 1 1 x2 0 3 1 5 x3 Chapter I: Matrices I – 1.35 x1 x2 x3 ∴ 4 8 4 x1 x2 x3 or 1 2 1 1 ∴ X3 2 1 Note Since the eigenvalues of A are distinct, the eigenvectors X1, X2, X3 are linearly independent, as can be seen from the fact that the equation k1X1 + k2X2 + k3X3 = 0 is satisﬁed only when k1 = k2 = k3 = 0. Example 1.3 Find the eigenvalues and eigenvectors of the matrix 0 1 1 A 1 0 1 1 1 0 The characteristic equation is given by λ3 – D1λ2 + D2λ – D3 = 0, where D1 = the sum of the ﬁrst order minors of A that lie along the main diagonal of A =0+0+0 =0 D2 = the sum of the second order minors of A whose principal diagonals lie along the principal diagonal of A. 0 1 0 1 0 1 1 0 1 0 1 0 =–3 D3= |A| = 2 Thus the characteristic equation of A is λ3 – 3λ – 2 = 0 i.e. (λ + l)2 (λ – 2) = 0 ∴ The eigenvalues of A are λ = –1, –1, 2. Case (i) λ = –1. The eigenvector is given by 1 1 x1 1 1 x2 0 1 1 x3 All the three equations reduce to one and the same equation x1 + x2 + x3 = 0. There is one equation in three unknowns. ∴ Two of the unknowns, say, x1 and x2 are to be treated as free variables (parameters). Taking x1 = 1 and x2 = 0, we get x3 = –1 and taking x1 = 0 and x2= 1, we get x3= – l I – 1.36 Part I: Mathematics I 1 0 X1 0 and X2 1 1 1 Case (ii) λ = 2. The eigenvector is given by 2 1 1 x1 1 2 1 x2 0 1 1 2 x3 Values of x1, x2, x3 are proportional to the co-factors of elements in the ﬁrst row. i.e. x1 x2 x3 3 3 3 x1 x2 x3 or 1 1 1 1 ∴ x3 1 1 Note Though two of the eigenvalues are equal, the eigenvectors X1, X2, X3 are found to be linearly independent. Example 1.4 Find the eigenvalues and eigenvectors of the matrix 2 2 2 A 1 1 1 1 3 1 The characteristic equation of A is 2 2 2 1 1 1 0 1 3 1 i.e. (2 – λ)(λ2 – 4) + 2(– l – λ – l) + 2(3 – l + λ) = 0 i.e. (2 – λ)(λ – 2)(λ + 2) = 0 ∴ The eigenvalues of A are λ = –2, 2, 2. Case (i) λ=–2 The eigenvector is given by 4 2 2 x1 1 3 1 x2 0 1 3 1 x3 x1 x2 x3 ∴ (by taking the co-factors of elements of the third row) 8 2 14 Chapter I: Matrices I – 1.37 x1 x2 x3 i.e. 4 1 7 4 ∴ X1 1 Case (ii) λ = 2. 7 The eigenvector is given by 0 2 2 x1 1 1 1 x2 0 1 3 3 x3 x1 x2 x3 x1 x2 x3 ∴ or 0 4 4 0 1 1 0 ∴ X2 X3 1 1 Note Two eigenvalues are equal and the eigenvectors are linearly dependent. Example 1.5 Find the eigenvalues and eigenvectors of the matrix 11 4 7 A 7 2 5 10 4 6 Can you guess the nature of A from the eigenvalues? Verify your answer. The characteristic equation of A is 11 4 7 7 2 5 0 10 4 6 i.e. (11 – λ)( λ2 + 8λ – 8) + 4(8 – 7λ)–7(10λ – 8) = 0 i.e. λ3 – 3λ2 + 2λ = 0 ∴ The eigenvalues of A are λ = 0, 1, 2. Case (i) λ = 0. 11 4 7 x1 The eigenvector is given by 7 2 5 x2 0 10 4 6 x3 ∴ x1 x2 x3 8 8 8 I – 1.38 Part I: Mathematics I or x1 x2 x3 1 1 1 1 ∴ X1 1 1 Case (ii) λ = 1. 10 4 7 x1 The eigenvector is given by 7 3 5 x2 0 10 4 7 x3 x1 x2 x3 ∴ 1 1 2 1 ∴ X2 1 2 Case (iii) λ = 2. 9 4 7 x1 The eigenvector is given by 7 4 5 x2 0 10 4 8 x3 x1 x2 x3 ∴ 12 6 12 x1 x2 x3 or 2 1 2 2 ∴ X3 1 2 Since one of the eigenvalues of A is zero, product of the eigenvalues = |A| = 0, i.e. A is non-singular. It is veriﬁed below: 11 4 7 7 2 5 11 12 20 4 42 50 7 28 20 0. 10 4 6 Example 1.6 Verify that the sum of the eigenvalues of A equals the trace of A and that their product equals |A|, for the matrix 1 0 0 A 0 3 1 0 1 3 Chapter I: Matrices I – 1.39 The characteristic equation of A is 1 0 0 0 3 1 0 0 1 3 i.e. (l – λ)(λ2 – 6λ + 8) = 0 ∴ The

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