Steven J. Leon University of Massachusetts, Dartmouth
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1 INSTRUCTOR S SOLUTIONS MANUAL LINEAR ALGEBRA WITH APPLICATIONS NINTH EDITION Steven J. Leon University of Massachusetts, Dartmouth Boston Columbus Indianapolis New York San Francisco Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
2 The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the author. Copyright 205, 200, 2006 Pearson Education, Inc. Publishing as Pearson, 75 Arlington Street, Boston, MA 026. All rights reserved. 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 prior written permission of the publisher. Printed in the United States of America. ISBN-3: ISBN-0: X OPM
3 Contents Preface v Matrices and Systems of Equations Systems of Linear Equations 2 Row Echelon Form 2 3 Matrix Arithmetic 3 4 Matrix Algebra 6 5 Elementary Matrices 2 6 Partitioned Matrices 7 MATLAB Exercises 20 Chapter Test A 22 Chapter Test B 24 2 Determinants 27 The Determinant of a Matrix 27 2 Properties of Determinants 30 3 Additional Topics and Applications 33 MATLAB Exercises 35 Chapter Test A 35 Chapter Test B 36 3 Vector Spaces 38 Definition and Examples 38 2 Subspaces 42 3 Linear Independence 47 4 Basis and Dimension 50 5 Change of Basis 52 6 Row Space and Column Space 52 MATLAB Exercises 59 Chapter Test A 60 Chapter Test B 62 4 Linear Transformations 66 Definition and Examples 66 2 Matrix Representations of Linear Transformations 69 3 Similarity 7 MATLAB Exercise 72 iii
4 iv Contents Chapter Test A 73 Chapter Test B 74 5 Orthogonality 76 The Scalar product in R n 76 2 Orthogonal Subspaces 78 3 Least Squares Problems 8 4 Inner Product Spaces 85 5 Orthonormal Sets 90 6 The Gram-Schmidt Process 98 7 Orthogonal Polynomials 00 MATLAB Exercises 03 Chapter Test A 04 Chapter Test B 05 6 Eigenvalues 09 Eigenvalues and Eigenvectors 09 2 Systems of Linear Differential Equations 4 3 Diagonalization 5 4 Hermitian Matrices 23 5 Singular Value Decomposition 30 6 Quadratic Forms 32 7 Positive Definite Matrices 35 8 Nonnegative Matrices 38 MATLAB Exercises 40 Chapter Test A 44 Chapter Test B 45 7 Numerical Linear Algebra 49 Floating-Point Numbers 49 2 Gaussian Elimination 50 3 Pivoting Strategies 5 4 Matrix Norms and Condition Numbers 52 5 Orthogonal Transformations 62 6 The Eigenvalue Problem 64 7 Least Squares Problems 68 MATLAB Exercises 7 Chapter Test A 72 Chapter Test B 73
5 Preface This solutions manual is designed to accompany the ninth edition of Linear Algebra with Applications by Steven J. Leon. The answers in this manual supplement those given in the answer key of the textbook. In addition, this manual contains the complete solutions to all of the nonroutine exercises in the book. At the end of each chapter of the textbook there are two chapter tests (A and B) and a section of computer exercises to be solved using MATLAB. The questions in each Chapter Test A are to be answered as either true or false. Although the true-false answers are given in the Answer Section of the textbook, students are required to explain or prove their answers. This manual includes explanations, proofs, and counterexamples for all Chapter Test A questions. The chapter tests labeled B contain problems similar to the exercises in the chapter. The answers to these problems are not given in the Answers to Selected Exercises Section of the textbook; however, they are provided in this manual. Complete solutions are given for all of the nonroutine Chapter Test B exercises. In the MATLAB exercises. most of the computations are straightforward. Consequently, they have not been included in this solutions manual. On the other hand, the text also includes questions related to the computations. The purpose of the questions is to emphasize the significance of the computations. The solutions manual does provide the answers to most of these questions. There are some questions for which it is not possible to provide a single answer. For example, some exercises involve randomly generated matrices. In these cases, the answers may depend on the particular random matrices that were generated. Steven J. Leon sleon@umassd.edu v
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7 Chapter Matrices and Systems of Equations SYSTEMS OF LINEAR EQUATIONS (d) (a) 3x + 2x 2 = 8 x + 5x 2 = 7 (b) 5x 2x 2 + x 3 = 3 2x + 3x 2 4x 3 = 0 (c) 2x + x 2 + 4x 3 = 4x 2x 2 + 3x 3 = 4 5x + 2x 2 + 6x 2 =
8 2 Chapter Matrices and Systems of Equations (d) 4x 3x 2 + x 3 + 2x 4 = 4 3x + x 2 5x 3 + 6x 4 = 5 x + x 2 + 2x 3 + 4x 4 = 8 5x + x 2 + 3x 3 2x 4 = 7 9. Given the system m x + x 2 = b m 2 x + x 2 = b 2 one can eliminate the variable x 2 by subtracting the first row from the second. One then obtains the equivalent system m x + x 2 = b (m m 2 )x = b 2 b (a) If m m 2, then one can solve the second equation for x x = b 2 b m m 2 One can then plug this value of x into the first equation and solve for x 2. Thus, if m m 2, there will be a unique ordered pair (x, x 2 ) that satisfies the two equations. (b) If m = m 2, then the x term drops out in the second equation 0 = b 2 b This is possible if and only if b = b 2. (c) If m m 2, then the two equations represent lines in the plane with different slopes. Two nonparallel lines intersect in a point. That point will be the unique solution to the system. If m = m 2 and b = b 2, then both equations represent the same line and consequently every point on that line will satisfy both equations. If m = m 2 and b b 2, then the equations represent parallel lines. Since parallel lines do not intersect, there is no point on both lines and hence no solution to the system. 0. The system must be consistent since (0, 0) is a solution.. A linear equation in 3 unknowns represents a plane in three space. The solution set to a 3 3 linear system would be the set of all points that lie on all three planes. If the planes are parallel or one plane is parallel to the line of intersection of the other two, then the solution set will be empty. The three equations could represent the same plane or the three planes could all intersect in a line. In either case the solution set will contain infinitely many points. If the three planes intersect in a point, then the solution set will contain only that point. 2 ROW ECHELON FORM 2. (b) The system is consistent with a unique solution (4, ). 4. (b) x and x 3 are lead variables and x 2 is a free variable. (d) x and x 3 are lead variables and x 2 and x 4 are free variables. (f) x 2 and x 3 are lead variables and x is a free variable. 5. (l) The solution is (0,.5, 3.5). 6. (c) The solution set consists of all ordered triples of the form (0, α, α). 7. A homogeneous linear equation in 3 unknowns corresponds to a plane that passes through the origin in 3-space. Two such equations would correspond to two planes through the origin. If one equation is a multiple of the other, then both represent the same plane through the origin and every point on that plane will be a solution to the system. If one equation is not a multiple of the other, then we have two distinct planes that intersect in a line through the
9 Section 3 Matrix Arithmetic 3 origin. Every point on the line of intersection will be a solution to the linear system. So in either case the system must have infinitely many solutions. In the case of a nonhomogeneous 2 3 linear system, the equations correspond to planes that do not both pass through the origin. If one equation is a multiple of the other, then both represent the same plane and there are infinitely many solutions. If the equations represent planes that are parallel, then they do not intersect and hence the system will not have any solutions. If the equations represent distinct planes that are not parallel, then they must intersect in a line and hence there will be infinitely many solutions. So the only possibilities for a nonhomogeneous 2 3 linear system are 0 or infinitely many solutions. 9. (a) Since the system is homogeneous it must be consistent. 3. A homogeneous system is always consistent since it has the trivial solution (0,..., 0). If the reduced row echelon form of the coefficient matrix involves free variables, then there will be infinitely many solutions. If there are no free variables, then the trivial solution will be the only solution. 4. A nonhomogeneous system could be inconsistent in which case there would be no solutions. If the system is consistent and underdetermined, then there will be free variables and this would imply that we will have infinitely many solutions. 6. At each intersection, the number of vehicles entering must equal the number of vehicles leaving in order for the traffic to flow. This condition leads to the following system of equations x + a = x 2 + b x 2 + a 2 = x 3 + b 2 x 3 + a 3 = x 4 + b 3 If we add all four equations, we get x 4 + a 4 = x + b 4 and hence x + x 2 + x 3 + x 4 + a + a 2 + a 3 + a 4 = x + x 2 + x 3 + x 4 + b + b 2 + b 3 + b 4 7. If (c, c 2 ) is a solution, then a + a 2 + a 3 + a 4 = b + b 2 + b 3 + b 4 a c + a 2 c 2 = 0 a 2 c + a 22 c 2 = 0 Multiplying both equations through by α, one obtains a (αc ) + a 2 (αc 2 ) = α 0 = 0 a 2 (αc ) + a 22 (αc 2 ) = α 0 = 0 Thus (αc, αc 2 ) is also a solution. 8. (a) If x 4 = 0, then x, x 2, and x 3 will all be 0. Thus if no glucose is produced, then there is no reaction. (0, 0, 0, 0) is the trivial solution in the sense that if there are no molecules of carbon dioxide and water, then there will be no reaction. (b) If we choose another value of x 4, say x 4 = 2, then we end up with solution x = 2, x 2 = 2, x 3 = 2, x 4 = 2. Note the ratios are still 6:6:6:. 3 MATRIX ARITHMETIC. (e)
10 4 Chapter Matrices and Systems of Equations (g) (d) (a) 5A = A + 3A = (b) 6A = = (2A) = = (c) A T = (A T ) T = 3 2 T 3 4 = 4 7 = A (a) A + B = = B + A 0 5 (b) 3(A + B) = = A + 3B = = (c) (A + B) T = T 5 0 = A T + B T = = (a) 3(AB) = 3 (3A)B = A(3B) = = = =
11 Section 3 Matrix Arithmetic T = (b) (AB) T = B T A T = = (a) (A + B) + C = = A + (B + C) = = (b) (AB)C = = A(BC) = = (c) A(B + C) = = AB + AC = = (d) (A + B)C = = AC + BC = = (b) x = (2, ) T is a solution since b = 2a +a 2. There are no other solutions since the echelon form of A is strictly triangular. (c) The solution to Ax = c is x = ( 5 2, 4 )T. Therefore c = 5 2 a 4 a 2.. The given information implies that 0 x = and x 2 = 0 are both solutions to the system. So the system is consistent and since there is more than one solution, the row echelon form of A must involve a free variable. A consistent system with a free variable has infinitely many solutions. 2. The system is consistent since x = (,,, ) T is a solution. The system can have at most 3 lead variables since A only has 3 rows. Therefore, there must be at least one free variable. A consistent system with a free variable has infinitely many solutions. 3. (a) It follows from the reduced row echelon form that the free variables are x 2, x 4, x 5. If we set x 2 = a, x 4 = b, x 5 = c, then x x 3 = 2 2a 3b c = 5 2b 4c and hence the solution consists of all vectors of the form x = ( 2 2a 3b c, a, 5 2b 4c, b, c) T (b) If we set the free variables equal to 0, then x 0 = ( 2, 0, 5, 0, 0) T is a solution to Ax = b and hence b = Ax 0 = 2a + 5a 3 = (8, 7,, 7) T
12 6 Chapter Matrices and Systems of Equations 4. If w 3 is the weight given to professional activities, then the weights for research and teaching should be w = 3w 3 and w 2 = 2w 3. Note that.5w 2 = 3w 3 = w, so the weight given to research is.5 times the weight given to teaching. Since the weights must all add up to, we have = w + w 2 + w 3 = 3w 3 + 2w 3 + w 3 = 6w 3 and hence it follows that w 3 = 6, w 2 = 3, w = 2. If C is the matrix in the example problem from the Analytic Hierarchy Process Application, then the rating vector r is computed by multiplying C times the weight vector w. r = Cw = = 5. A T is an n m matrix. Since A T has m columns and A has m rows, the multiplication A T A is possible. The multiplication AA T is possible since A has n columns and A T has n rows. 6. If A is skew-symmetric, then A T = A. Since the (j, j) entry of A T is a jj and the (j, j) entry of A is a jj, it follows that a jj = a jj for each j and hence the diagonal entries of A must all be The search vector is x = (, 0,, 0,, 0) T. The search result is given by the vector y = A T x = (, 2, 2,,, 2, ) T The ith entry of y is equal to the number of search words in the title of the ith book. 8. If α = a 2 /a, then 0 a a 2 = a a 2 = a a 2 α 0 b αa αa 2 + b a 2 αa 2 + b The product will equal A provided αa 2 + b = a Thus we must choose b = a 22 αa 2 = a 22 a 2a 2 a 4 MATRIX ALGEBRA. (a) (A + B) 2 = (A + B)(A + B) = (A + B)A + (A + B)B = A 2 + BA + AB + B 2 For real numbers, ab + ba = 2ab; however, with matrices AB + BA is generally not equal to 2AB. (b) (A + B)(A B) = (A + B)(A B) = (A + B)A (A + B)B = A 2 + BA AB B 2 For real numbers, ab ba = 0; however, with matrices AB BA is generally not equal to O.
13 Section 4 Matrix Algebra 7 2. If we replace a by A and b by the identity matrix, I, then both rules will work, since and (A + I) 2 = A 2 + IA + AI + B 2 = A 2 + AI + AI + B 2 = A 2 + 2AI + B 2 (A + I)(A I) = A 2 + IA AI I 2 = A 2 + A A I 2 = A 2 I 2 3. There are many possible choices for A and B. For example, one could choose A = 0 and B = More generally if A = a ca b cb B = db eb da ea then AB = O for any choice of the scalars a, b, c, d, e. 4. To construct nonzero matrices A, B, C with the desired properties, first find nonzero matrices C and D such that DC = O (see Exercise 3). Next, for any nonzero matrix A, set B = A+D. It follows that BC = (A + D)C = AC + DC = AC + O = AC 5. A 2 2 symmetric matrix is one of the form A = a b Thus b c A 2 = a2 + b 2 ab + bc ab + bc b 2 + c 2 If A 2 = O, then its diagonal entries must be 0. a 2 + b 2 = 0 and b 2 + c 2 = 0 Thus a = b = c = 0 and hence A = O. 6. Let D = (AB)C = a b + a 2 b 2 a b 2 + a 2 b 22 a 2 b + a 22 b 2 a 2 b 2 + a 22 b 22 It follows that If we set E = A(BC) = then it follows that d = (a b + a 2 b 2 )c + (a b 2 + a 2 b 22 )c 2 = a b c + a 2 b 2 c + a b 2 c 2 + a 2 b 22 c 2 d 2 = (a b + a 2 b 2 )c 2 + (a b 2 + a 2 b 22 )c 22 = a b c 2 + a 2 b 2 c 2 + a b 2 c 22 + a 2 b 22 c 22 d 2 = (a 2 b + a 22 b 2 )c + (a 2 b 2 + a 22 b 22 )c 2 = a 2 b c + a 22 b 2 c + a 2 b 2 c 2 + a 22 b 22 c 2 d 22 = (a 2 b + a 22 b 2 )c 2 + (a 2 b 2 + a 22 b 22 )c 22 = a 2 b c 2 + a 22 b 2 c 2 + a 2 b 2 c 22 + a 22 b 22 c 22 a a 2 a 2 a 22 c c 2 c 2 c 22 b c + b 2 c 2 b c 2 + b 2 c 22 b 2 c + b 22 c 2 b 2 c 2 + b 22 c 22 e = a (b c + b 2 c 2 ) + a 2 (b 2 c + b 22 c 2 ) = a b c + a b 2 c 2 + a 2 b 2 c + a 2 b 22 c 2
14 8 Chapter Matrices and Systems of Equations 9. Thus and hence and A 4 = O. If n > 4, then e 2 = a (b c 2 + b 2 c 22 ) + a 2 (b 2 c 2 + b 22 c 22 ) = a b c 2 + a b 2 c 22 + a 2 b 2 c 2 + a 2 b 22 c 22 e 2 = a 2 (b c + b 2 c 2 ) + a 22 (b 2 c + b 22 c 2 ) = a 2 b c + a 2 b 2 c 2 + a 22 b 2 c + a 22 b 22 c 2 e 22 = a 2 (b c 2 + b 2 c 22 ) + a 22 (b 2 c 2 + b 22 c 22 ) = a 2 b c 2 + a 2 b 2 c 22 + a 22 b 2 c 2 + a 22 b 22 c 22 d = e d 2 = e 2 d 2 = e 2 d 22 = e 22 (AB)C = D = E = A(BC) A = (a) The matrix C is symmetric since (b) The matrix D is symmetric since A = A n = A n 4 A 4 = A n 4 O = O (c) The matrix E = AB is not symmetric since and in general, AB BA. (d) The matrix F is symmetric since (e) The matrix G is symmetric since C T = (A + B) T = A T + B T = A + B = C D T = (AA) T = A T A T = A 2 = D E T = (AB) T = B T A T = BA F T = (ABA) T = A T B T A T = ABA = F G T = (AB + BA) T = (AB) T + (BA) T = B T A T + A T B T = BA + AB = G (f) The matrix H is not symmetric since H T = (AB BA) T = (AB) T (BA) T = B T A T A T B T = BA AB = H. (a) The matrix A is symmetric since (b) The matrix B is not symmetric since (c) The matrix D is symmetric since (d) The matrix E is symmetric since A T = (C + C T ) T = C T + (C T ) T = C T + C = A B T = (C C T ) T = C T (C T ) T = C T C = B E T A T = (C T C) T = C T (C T ) T = C T C = D = (C T C CC T ) T = (C T C) T (CC T ) T = C T (C T ) T (C T ) T C T = C T C CC T = E
15 Section 4 Matrix Algebra 9 (e) The matrix F is symmetric since F T = ((I + C)(I + C T )) T = (I + C T ) T (I + C) T = (I + C)(I + C T ) = F (e) The matrix G is not symmetric. F T F = (I + C)(I C T ) = I + C C T CC T = ((I + C)(I C T )) T = (I C T ) T (I + C) T = (I C)(I + C T ) = I C + C T CC T F and F T are not the same. The two middle terms C C T and C + C T do not agree. 2. If d = a a 22 a 2 a 2 0, then a 22 a 2 a a a 22 a 2 a 2 0 a 2 d = d a 2 a a 2 a 22 0 a a 22 a 2 a 2 = I d a [ a 2 a ] a a 22 a 2 a a 2 d = a 2 a 22 d a 2 a 0 a a 22 a 2 a 2 = I d Therefore a 22 a 2 = A d a 2 a 3. (b) If A were nonsingular and AB = A, then it would follow that A AB = A A and hence that B = I. So if B I, then A must be singular. 5. Since A A = AA = I it follows from the definition that A is nonsingular and its inverse is A. 6. Since it follows that A T (A ) T = (A A) T = I (A ) T A T = (AA ) T = I (A ) T = (A T ) 7. If Ax = Ay and x y, then A must be singular, for if A were nonsingular, then we could multiply by A and get 8. For m =, A Ax = A Ay x = y (A ) = A = (A ) Assume the result holds in the case m = k, that is, It follows that and (A k ) = (A ) k (A ) k+ A k+ = A (A ) k A k A = A A = I A k+ (A ) k+ = AA k (A ) k A = AA = I
16 0 Chapter Matrices and Systems of Equations Therefore (A ) k+ = (A k+ ) and the result follows by mathematical induction. 9. If A 2 = O, then (I + A)(I A) = I + A A + A 2 = I and (I A)(I + A) = I A + A + A 2 = I Therefore I A is nonsingular and (I A) = I + A. 20. If A k+ = O, then and (I + A + + A k )(I A) = (I + A + + A k ) (A + A A k+ ) = I A k+ = I (I A)(I + A + + A k ) = (I + A + + A k ) (A + A A k+ ) = I A k+ = I Therefore I A is nonsingular and (I A) = I + A + A A k. 2. Since R T cos θ sin θ R = cos θ sin θ = 0 sin θ cos θ sin θ cos θ and RR T = cos θ sin θ it follows that R is nonsingular and R = R T G 2 = H 2 = (I 2uu T ) 2 sin θ cos θ sin θ = 0 cos θ sin θ cos θ 0 cos2 θ + sin 2 θ 0 0 cos 2 θ + sin 2 θ = I 4uu T + 4uu T uu T = I 4uu T + 4u(u T u)u T = I = I 4uu T + 4uu T = I (since u T u = ) 24. In each case, if you square the given matrix, you will end up with the same matrix. 25. (a) If A 2 = A, then (I A) 2 = I 2A + A 2 = I 2A + A = I A (b) If A 2 = A, then (I 2 A)(I + A) = I 2 A + A 2 A2 = I 2 A + A 2 A = I and (I + A)(I 2 A) = I + A 2 A 2 A2 = I + A 2 A 2 A = I 26. (a) Therefore I + A is nonsingular and (I + A) = I 2 A. D 2 = d d d 2 nn
17 Section 4 Matrix Algebra Since each diagonal entry of D is equal to either 0 or, it follows that d 2 jj = d jj, for j =,..., n and hence D 2 = D. (b) If A = XDX, then A 2 = (XDX )(XDX ) = XD(X X)DX = XDX = A 27. If A is an involution, then A 2 = I and it follows that B 2 C 2 = 4 (I + A)2 = 4 (I + 2A + A2 ) = 4 (2I + 2A) = (I + A) = B 2 = 4 (I A)2 = 4 (I 2A + A2 ) = 4 (2I 2A) = (I A) = C 2 So B and C are both idempotent. BC = 4 (I + A)(I A) = 4 (I + A A A2 ) = (I + A A I) = O (A T A) T = A T (A T ) T = A T A (AA T ) T = (A T ) T A T = AA T 29. Let A and B be symmetric n n matrices. If (AB) T = AB, then 30. (a) Conversely, if BA = AB, then BA = B T A T = (AB) T = AB (AB) T = B T A T = BA = AB B T C T = (A + A T ) T = A T + (A T ) T = A T + A = B = (A A T ) T = A T (A T ) T = A T A = C (b) A = 2 (A + AT ) + 2 (A AT ) 34. False. For example, if A = 2 3, B = 4, x = then Ax = Bx = 5 5 however, A B. 35. False. For example, if A = 0 and B = then it is easy to see that both A and B must be singular, however, A + B = I, which is nonsingular. 36. True. If A and B are nonsingular, then their product AB must also be nonsingular. Using the result from Exercise 23, we have that (AB) T is nonsingular and ((AB) T ) = ((AB) ) T. It follows then that ((AB) T ) = ((AB) ) T = (B A ) T = (A ) T (B ) T
18 2 Chapter Matrices and Systems of Equations 5 ELEMENTARY MATRICES 2. (a) 0, type I 0 (b) The given matrix is not an elementary matrix. Its inverse is given by (c) (d) 5. (c) Since / , type III, type II C = F B = F EA where F and E are elementary matrices, it follows that C is row equivalent to A (b) E = 3 0, E 2 = 0 0, E 3 = The product L = E E 2 E 3 is lower triangular. 0 0 L = A can be reduced to the identity matrix using three row operations The elementary matrices corresponding to the three row operations are E = 0, E 3 2 =, E 0 3 = So and hence and A = E 3 E 2 E. 8. (b) (d) (a) E 3 E 2 E A = I A = E E 3 E 3 = =
19 (e) (b) XA + B = C X = (C B)A = (d) XA + C = X XA XI = C X(A I) = C X = C(A I) = = Section 5 Elementary Matrices (a) If E is an elementary matrix of type I or type II, then E is symmetric. Thus E T = E is an elementary matrix of the same type. If E is the elementary matrix of type III formed by adding α times the ith row of the identity matrix to the jth row, then E T is the elementary matrix of type III formed from the identity matrix by adding α times the jth row to the ith row. (b) In general, the product of two elementary matrices will not be an elementary matrix. Generally, the product of two elementary matrices will be a matrix formed from the identity matrix by the performance of two row operations. For example, if E = 2 0 and E 2 = then E and E 2 are elementary matrices, but 0 0 E E 2 = is not an elementary matrix. 4. If T = UR, then Since U and R are upper triangular If i > j, then t ij = n u ik r kj k= u i = u i2 = = u i,i = 0 r j+,j = r j+2,j = r nj = 0 t ij = = = 0 j u ik r kj + k= j 0 r kj + k= n k=j+ n k=j+ u ik r kj u ik 0
20 4 Chapter Matrices and Systems of Equations Therefore T is upper triangular. If i = j, then Therefore t jj = t ij = 5. If we set x = (2, 4) T, then = i u ik r kj + u jj r jj + k= i 0 r kj + u jj r jj + k= = u jj r jj t jj = u jj r jj n k=j+ n k=j+ j =,..., n Ax = 2a + a 2 4a 3 = 0 u ik r kj Thus x is a nonzero solution to the system Ax = 0. But if a homogeneous system has a nonzero solution, then it must have infinitely many solutions. In particular, if c is any scalar, then cx is also a solution to the system since A(cx) = cax = c0 = 0 Since Ax = 0 and x 0, it follows that the matrix A must be singular. (See Theorem.5.2) 6. If a = 3a 2 2a 3, then a 3a 2 + 2a 3 = 0 Therefore x = (, 3, 2) T is a nontrivial solution to Ax = 0. It follows from Theorem.5.2 that A must be singular. 7. If x 0 0 and Ax 0 = Bx 0, then Cx 0 = 0 and it follows from Theorem.5.2 that C must be singular. 8. If B is singular, then it follows from Theorem.5.2 that there exists a nonzero vector x such that Bx = 0. If C = AB, then Cx = ABx = A0 = 0 Thus, by Theorem.5.2, C must also be singular. 9. (a) If U is upper triangular with nonzero diagonal entries, then using row operation II, U can be transformed into an upper triangular matrix with s on the diagonal. Row operation III can then be used to eliminate all of the entries above the diagonal. Thus, U is row equivalent to I and hence is nonsingular. (b) The same row operations that were used to reduce U to the identity matrix will transform I into U. Row operation II applied to I will just change the values of the diagonal entries. When the row operation III steps referred to in part (a) are applied to a diagonal matrix, the entries above the diagonal are filled in. The resulting matrix, U, will be upper triangular. 20. Since A is nonsingular it is row equivalent to I. Hence, there exist elementary matrices E, E 2,..., E k such that E k E A = I It follows that and A = E k E E k E B = A B = C The same row operations that reduce A to I, will transform B to C. Therefore, the reduced row echelon form of (A B) will be (I C). u ik 0
21 Section 5 Elementary Matrices 5 2. (a) If the diagonal entries of D are α, α 2,..., α n and the diagonal entries of D 2 are β, β 2,..., β n, then D D 2 will be a diagonal matrix with diagonal entries α β,..., α n β n and D 2 D will be a diagonal matrix with diagonal entries β α, β 2 α 2,..., β n α n. Since the two have the same diagonal entries, it follows that D D 2 = D 2 D. (b) AB = A(a 0 I + a A + + a k A k ) = a 0 A + a A a k A k+ = (a 0 I + a A + + a k A k )A = BA 22. If A is symmetric and nonsingular, then (A ) T = (A ) T (AA ) = ((A ) T A T )A = A 23. If A is row equivalent to B, then there exist elementary matrices E, E 2,..., E k such that Each of the E i s is invertible and E i A = E k E k E B is also an elementary matrix (Theorem.4.). Thus B = E E 2 E A and hence B is row equivalent to A. 24. (a) If A is row equivalent to B, then there exist elementary matrices E, E 2,..., E k such that A = E k E k E B Since B is row equivalent to C, there exist elementary matrices H, H 2,..., H j such that Thus B = H j H j H C A = E k E k E H j H j H C and hence A is row equivalent to C. (b) If A and B are nonsingular n n matrices, then A and B are row equivalent to I. Since A is row equivalent to I and I is row equivalent to B, it follows from part (a) that A is row equivalent to B. 25. If U is any row echelon form of A, then A can be reduced to U using row operations, so A is row equivalent to U. If B is row equivalent to A, then it follows from the result in Exercise 24(a) that B is row equivalent to U. 26. If B is row equivalent to A, then there exist elementary matrices E, E 2,..., E k such that B = E k E k E A Let M = E k E k E. The matrix M is nonsingular since each of the E i s is nonsingular. Conversely, suppose there exists a nonsingular matrix M such that B = MA. Since M is nonsingular, it is row equivalent to I. Thus, there exist elementary matrices E, E 2,..., E k such that M = E k E k E I It follows that Therefore, B is row equivalent to A. B = MA = E k E k E A k
22 6 Chapter Matrices and Systems of Equations 27. If A is nonsingular, then A is row equivalent to I. If B is row equivalent to A, then using the result from Exercise 24(a), we can conclude that B is row equivalent to I. Therefore, B must be nonsingular. So it is not possible for B to be singular and also be row equivalent to a nonsingular matrix. 28. (a) The system V c = y is given by x x 2 x n x 2 x 2 2 x n 2.. x n+ x 2 n+ x n n+ Comparing the ith row of each side, we have c c 2.. c n+ = y y 2. y n+ c + c 2 x i + + c n+ x n i = y i Thus p(x i ) = y i i =, 2,..., n + (b) If x, x 2,..., x n+ are distinct and V c = 0, then we can apply part (a) with y = 0. Thus if p(x) = c + c 2 x + + c n+ x n, then p(x i ) = 0 i =, 2,..., n + The polynomial p(x) has n + roots. Since the degree of p(x) is less than n +, p(x) must be the zero polynomial. Hence c = c 2 = = c n+ = 0 Since the system V c = 0 has only the trivial solution, the matrix V must be nonsingular. 29. True. If A is row equivalent to I, then A is nonsingular, so if AB = AC, then we can multiply both sides of this equation by A. A AB = A AC B = C 30. True. If E and F are elementary matrices, then they are both nonsingular and the product of two nonsingular matrices is a nonsingular matrix. Indeed, G = F E. 3. True. If a + a 2 = a 3 + 2a 4, then a + a 2 a 3 2a 4 = 0 If we let x = (,,, 2) T, then x is a solution to Ax = 0. Since x 0 the matrix A must be singular. 32. False. Let I be the 2 2 identity matrix and let A = I, B = I, and C = Since B and C are nonsingular, they are both row equivalent to A; however, B + C = is singular, so it cannot be row equivalent to A.
23 Section 6 Partitioned Matrices 7 6 PARTITIONED MATRICES 2. B = A T A = 5. (a) a T a T 2.. a T n 2 2 (c) Let (a, a 2,..., a n ) = a T a a T a 2 a T a n a T 2 a a T 2 a 2 a T 2 a n.. a T n a a T n a 2 a T n a n + ( 2 3) = A = A 2 = (a) A 2 = (0 0) A 22 = ( 0) The block multiplication is performed as follows: A A 2 AT A T 2 A 2 A 22 A T 2 A T = A A T + A 2 A T 2 A A T 2 + A 2 A T A 2 A T + A 22 A T 2 A 2 A T 2 + A 22 A T = XY T = x y T + x 2 y T 2 + x 3 y T 3 = = (b) Since y i x T i = (x i y T i )T for j =, 2, 3, the outer product expansion of Y X T is just the transpose of the outer product expansion of XY T. Thus Y X T = y x T + y 2 x T 2 + y 3 x T 3 = It is possible to perform both block multiplications. To see this, suppose A is a k r matrix, A 2 is a k (n r) matrix, A 2 is an (m k) r matrix and A 22 is (m k) (n r). It is possible to perform the block multiplication of AA T since the matrix multiplications A A T, A A T 2, A 2 A T 2, A 2 A T 22, A 2 A T, A 2 A T 2, A 22 A T 2, A 22 A T 22 are all possible. It is possible to perform the block multiplication of A T A since the matrix multiplications A T A, A T A 2, A T 2A 2, A T 2A, A T 2A 2, A T 22A 2, A T 22A 22 are all possible. 8. AX = A(x, x 2,..., x r ) = (Ax, Ax 2,..., Ax r ) B = (b, b 2,..., b r ) AX = B if and only if the column vectors of AX and B are equal Ax j = b j j =,..., r 9. (a) Since D is a diagonal matrix, its jth column will have d jj in the jth row and the other entries will all be 0. Thus d j = d jj e j for j =,..., n.
24 8 Chapter Matrices and Systems of Equations (b) AD = A(d e, d 22 e 2,..., d nn e n ) = (d Ae, d 22 Ae 2,..., d nn Ae n ) = (d a, d 22 a 2,..., d nn a n ) 0. (a) (b) If we let X = UΣ, then and it follows that. If then Let UΣ = U U 2 Σ = U O Σ + U 2 O = U Σ X = U Σ = (σ u, σ 2 u 2,..., σ n u n ) A = UΣV T = XV T = σ u v T + σ 2 u 2 v T σ n u n v T n A C O A 22 A A 2 I = O A 22 O B = A A 2 + CA 22 = O C = A A 2A 22 A A A 2A 22 O A 22 A A 2 + CA 22 Since AB = BA = I, it follows that B = A. 2. Let 0 denote the zero vector in R n. If A is singular, then there exists a vector x 0 such that Ax = 0. If we set then Mx = A O x = x 0 O x = Ax + O0 = 0 B 0 Ox + B0 0 By Theorem.5.2, M must be singular. Similarly, if B is singular, then there exists a vector x 2 0 such that Bx 2 = 0. So if we set x = 0 x 2 I 5. then x is a nonzero vector and Mx is equal to the zero vector. A = O I, A 2 = I B, A 3 = B I B B I I and hence A + A 2 + A 3 = I + B 2I + B 2I + B I + B I 2B
25 Section 6 Partitioned Matrices 9 6. The block form of S is given by It follows that S MS = I O = I O = O B S = I O A AB I B A AB I B O BA 7. The block multiplication of the two factors yields I O A A 2 = B I O C A I O I O O ABA BA A I A A 2 BA BA 2 + C If we equate this matrix with the block form of A and solve for B and C, we get To check that this works note that and hence I O B I B = A 2 A and C = A 22 A 2 A A 2 BA = A 2 A A = A 2 BA 2 + C = A 2 A A 2 + A 22 A 2 A A 2 = A 22 A A 2 O C = A A 2 A 2 A In order for the block multiplication to work, we must have XB = S and Y M = T = A Since both B and M are nonsingular, we can satisfy these conditions by choosing X = SB and Y = T M. 9. (a) (b) b b 2... BC = b n (c) = Ax = (a, a 2,..., a n ) (c) It follows from parts (a) and (b) that b c b 2 c = cb. b n c x x 2.. x n = a (x ) + a 2 (x 2 ) + + a n (x n ) Ax = a (x ) + a 2 (x 2 ) + + a n (x n ) = x a + x 2 a x n a n
26 20 Chapter Matrices and Systems of Equations 20. If Ax = 0 for all x R n, then a j = Ae j = 0 for j =,..., n and hence A must be the zero matrix. 2. If Bx = Cx for all x R n then 22. (a) It follows from Exercise 20 that (b) If then A 0 A c T A (B C)x = 0 for all x R n c T I A a 0 T c T A a + β B C = O a β B = C x x n+ x x n+ = A 0 c T A = y = A a and z = A b ( c T y + β)x n+ = c T z + b n+ b b n+ A b c T A b + b n+ and x n+ = ct z + b n+ c T y + β x + x n+ A a = A b (β c T y 0) x = A b x n+ A a = z x n+ y MATLAB EXERCISES. In parts (a), (b), (c) it should turn out that A = A4 and A2 = A3. In part (d) A = A3 and A2 = A4. Exact equality will not occur in parts (c) and (d) because of roundoff error. 2. The solution x obtained using the \ operation will be more accurate and yield the smaller residual vector. The computation of x is also more efficient since the solution is computed using Gaussian elimination with partial pivoting and this involves less arithmetic than computing the inverse matrix and multiplying it times b. 3. (a) Since Ax = 0 and x 0, it follows from Theorem.5.2 that A is singular. (b) The columns of B are all multiples of x. Indeed, and hence (c) If D = B + C, then B = (x, 2x, 3x, 4x, 5x, 6x) AB = (Ax, 2Ax, 3Ax, 4Ax, 5Ax, 6Ax) = O AD = AB + AC = O + AC = AC 4. By construction, B is upper triangular whose diagonal entries are all equal to. Thus B is row equivalent to I and hence B is nonsingular. If one changes B by setting b 0, = /256 and computes Bx, the result is the zero vector. Since x 0, the matrix B must be singular.
27 MATLAB Exercises 2 5. (a) Since A is nonsingular, its reduced row echelon form is I. If E,..., E k are elementary matrices such that E k E A = I, then these same matrices can be used to transform (A b) to its reduced row echelon form U. It follows then that U = E k E (A b) = A (A b) = (I A b) Thus, the last column of U should be equal to the solution x of the system Ax = b. (b) After the third column of A is changed, the new matrix A is now singular. Examining the last row of the reduced row echelon form of the augmented matrix (A b), we see that the system is inconsistent. (c) The system Ax = c is consistent since y is a solution. There is a free variable x 3, so the system will have infinitely many solutions. (f) The vector v is a solution since Av = A(w + 3z) = Aw + 3Az = c For this solution, the free variable x 3 = v 3 = 3. To determine the general solution just set x = w + tz. This will give the solution corresponding to x 3 = t for any real number t. 6. (c) There will be no walks of even length from V i to V j whenever i + j is odd. (d) There will be no walks of length k from V i to V j whenever i + j + k is odd. (e) The conjecture is still valid for the graph containing the additional edges. (f) If the edge {V 6, V 8 } is included, then the conjecture is no longer valid. There is now a walk of length from V 6 to V 8 and i + j + k = is odd. 8. The change in part (b) should not have a significant effect on the survival potential for the turtles. The change in part (c) will effect the (2, 2) and (3, 2) of the Leslie matrix. The new values for these entries will be l 22 = and l 32 = With these values, the Leslie population model should predict that the survival period will double but the turtles will still eventually die out. 9. (b) x = c V x2. 0. (b) A 2k = I kb kb I This can be proved using mathematical induction. In the case k = A 2 = O I O I = I B I B I B B I If the result holds for k = m then A 2m+2 A 2m = I mb mb I = A 2 A 2m = I B I mb B I mb I I (m + )B = (m + )B I It follows by mathematical induction that the result holds for all positive integers k.
28 22 Chapter Matrices and Systems of Equations (b) A 2k+ = AA 2k = O I I I B kb kb = kb I I I (k + )B. (a) By construction, the entries of A were rounded to the nearest integer. The matrix B = A T A must also have integer entries and it is symmetric since B T = (A T A) T = A T (A T ) T = A T A = B (b) where It follows that Therefore LDL T = I O B O I E I O F O = B B E T EB E T + F EB ET E = B 2 B and F = B 22 B 2 B B 2 B E T EB I = B (B )T B2 T = B B B 2 = B 2 = B 2 B B = B 2 EB E T + F = B 2 E T + B 22 B 2 B B 2 = B 2 B B 2 + B 22 B 2 B B 2 = B 22 LDL T = B CHAPTER TEST A. The statement is false. If the row echelon form has free variables and the linear system is consistent, then there will be infinitely many solutions. However, it is possible to have an inconsistent system whose coefficient matrix will reduce to an echelon form with free variables. For example, if A = b = 0 0 then A involves one free variable, but the system Ax = b is inconsistent. 2. The statement is true since the zero vector will always be a solution. 3. The statement is true. A matrix A is nonsingular if and only if it is row equivalent to the I (the identity matrix). A will be row equivalent to I if and only if its reduced row echelon form is I. 4. The statement is true. If A is nonsingular, then A is row equivalent to I. So there exist elementary matrices E, E 2,..., E k, such that A = E k E k E I = E k E k E 5. The statement is false. For example, if A = I and B = I, the matrices A and B are both nonsingular, but A + B = O is singular.
29 Chapter Test A The statement is false. For example, if A is any matrix of the form A = cos θ sin θ sin θ cos θ Then A = A. 7. The statement is false. (A B) 2 = A 2 BA AB + B 2 A 2 2AB + B 2 since in general BA AB. For example, if A = and B = then however, (A B) 2 = 0 2 = 0 2 A 2 2AB + B 2 = = The statement is false. If A is nonsingular and AB = AC, then we can multiply both sides of the equation by A and conclude that B = C. However, if A is singular, then it is possible to have AB = AC and B C. For example, if A =, B =, C = then AB = = AC = 2 2 = The statement is false. In general, AB and BA are usually not equal, so it is possible for AB = O and BA to be a nonzero matrix. For example, if A = and B = then AB = 0 0 and BA = The statement is true. If x = (, 2, ) T, then x 0 and Ax = 0, so A must be singular.. The statement is true. If b = a + a 3 and x = (, 0, ) T, then Ax = x a + x 2 a 2 + x 3 a 3 = a + 0a 2 + a 3 = b So x is a solution to Ax = b. 2. The statement is true. If b = a + a 2 + a 3, then x = (,, ) T is a solution to Ax = b, since Ax = x a + x 2 a 2 + x 3 a 3 = a + a 2 + a 3 = b If a 2 = a 3, then we can also express b as a linear combination b = a + 0a 2 + 2a 3
30 24 Chapter Matrices and Systems of Equations Thus y = (, 0, 2) T is also a solution to the system. However, if there is more than one solution, then the reduced row echelon form of A must involve a free variable. A consistent system with a free variable must have infinitely many solutions. 3. The statement is true. An elementary matrix E of type I or type II is symmetric. So in either case we have E T = E is elementary. If E is an elementary matrix of type III formed from the identity matrix by adding a nonzero multiple c of row k to row j, then E T will be the elementary matrix of type III formed from the identity matrix by adding c times row j to row k. 4. The statement is false. An elementary matrix is a matrix that is constructed by performing exactly one elementary row operation on the identity matrix. The product of two elementary matrices will be a matrix formed by performing two elementary row operations on the identity matrix. For example, E = 2 0 and E 2 = are elementary matrices, however; 0 0 E E 2 = is not an elementary matrix. 5. The statement is true. The row vectors of A are x y T, x 2 y T,..., x n y T. Note, all of the row vectors are multiples of y T. Since x and y are nonzero vectors, at least one of these row vectors must be nonzero. However, if any nonzero row is picked as a pivot row, then since all of the other rows are multiples of the pivot row, they will all be eliminated in the first step of the reduction process. The resulting row echelon form will have exactly one nonzero row. CHAPTER TEST B The free variables are x 2 and x 4. If we set x 2 = a and x 4 = b, then x = 4 + a + 7b and x 3 = 3b and hence the solution set consists of all vectors of the form 4 + a + 7b a x = 3b b 2. (a) A linear equation in 3 unknowns corresponds to a plane in 3-space. (b) Given 2 equations in 3 unknowns, each equation corresponds to a plane. If one equation is a multiple of the other, then the equations represent the same plane and any point on the that plane will be a solution to the system. If the two planes are distinct, then they are either parallel or they intersect in a line. If they are parallel they do not intersect, so
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