Solution manual and test bank LInner algebra a modern INtroduction 4e

Complete Solutions Manual Linear Algebra A Modern Introduction FOURTH EDITION David Poole Trent University Prepared by Roger Lipsett Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States ISBN-13: 978-128586960-5 ISBN-10: 1-28586960-5 © 2015 Cengage Learning ALL RIGHTS RESERVED No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher except as may be permitted by the license terms below For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706 For permission to use 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helping to safeguard the integrity of the content contained in this Supplement We trust you find the Supplement a useful teaching tool Contents Vectors 1.1 The Geometry and Algebra of Vectors 1.2 Length and Angle: The Dot Product Exploration: Vectors and Geometry 1.3 Lines and Planes Exploration: The Cross Product 1.4 Applications Chapter Review 3 10 25 27 41 44 48 Systems of Linear Equations 2.1 Introduction to Systems of Linear Equations 2.2 Direct Methods for Solving Linear Systems Exploration: Lies My Computer Told Me Exploration: Partial Pivoting Exploration: An Introduction to the Analysis of Algorithms 2.3 Spanning Sets and Linear Independence 2.4 Applications 2.5 Iterative Methods for Solving Linear Systems Chapter Review 53 53 58 75 75 77 79 93 112 123 Matrices 3.1 Matrix Operations 3.2 Matrix Algebra 3.3 The Inverse of a Matrix 3.4 The LU Factorization 3.5 Subspaces, Basis, Dimension, and Rank 3.6 Introduction to Linear Transformations 3.7 Applications Chapter Review 129 129 138 150 164 176 192 209 230 Eigenvalues and Eigenvectors 4.1 Introduction to Eigenvalues and Eigenvectors 4.2 Determinants Exploration: Geometric Applications of Determinants 4.3 Eigenvalues and Eigenvectors of n × n Matrices 4.4 Similarity and Diagonalization 4.5 Iterative Methods for Computing Eigenvalues 4.6 Applications and the Perron-Frobenius Theorem Chapter Review 235 235 250 263 270 291 308 326 365 CONTENTS Orthogonality 5.1 Orthogonality in Rn 5.2 Orthogonal Complements and Orthogonal Projections 5.3 The Gram-Schmidt Process and the QR Factorization Exploration: The Modified QR Process Exploration: Approximating Eigenvalues with the QR Algorithm 5.4 Orthogonal Diagonalization of Symmetric Matrices 5.5 Applications Chapter Review 371 371 379 388 398 402 405 417 442 Vector Spaces 6.1 Vector Spaces and Subspaces 6.2 Linear Independence, Basis, and Dimension Exploration: Magic Squares 6.3 Change of Basis 6.4 Linear Transformations 6.5 The Kernel and Range of a Linear Transformation 6.6 The Matrix of a Linear Transformation Exploration: Tiles, Lattices, and the Crystallographic Restriction 6.7 Applications Chapter Review 451 451 463 477 480 491 498 507 525 527 531 Distance and Approximation 7.1 Inner Product Spaces Exploration: Vectors and Matrices with Complex Entries Exploration: Geometric Inequalities and Optimization Problems 7.2 Norms and Distance Functions 7.3 Least Squares Approximation 7.4 The Singular Value Decomposition 7.5 Applications Chapter Review 537 537 546 553 556 568 590 614 625 Codes 8.1 Code Vectors 8.2 Error-Correcting Codes 8.3 Dual Codes 8.4 Linear Codes 8.5 The Minimum Distance of 633 633 637 641 647 650 a Code Chapter Vectors 1.1 The Geometry and Algebra of Vectors 2, 2, 3 3, 1 3, 2 Since + = , −3 −3 −2 + = , −3 2 + = , −3 plotting those vectors gives c b a d 5 + = , −3 −2 −5 CHAPTER VECTORS c z b 0 y2 1 a x d Since the heads are all at (3, 2, 1), the tails are at       3 2 − 2 = 0 , 1       3 2 − 2 = 0 , 1       2 − −2 = 4 , 1 # » The four vectors AB are c d a b In standard position, the vectors are # » (a) AB = [4 − 1, − (−1)] = [3, 3] # » (b) AB = [2 − 0, −1 − (−2)] = [2, 1] # » (c) AB = 12 − 2, − 32 = − 32 , 32 # » (d) AB = 16 − 13 , 12 − 13 = − 16 , 16 a c b d 1       −1 2 − −1 = 3 −2 1.1 THE GEOMETRY AND ALGEBRA OF VECTORS Recall the notation that [a, b] denotes a move of a units horizontally and b units vertically Then during the first part of the walk, the hiker walks km north, so a = [0, 4] During the second part of the walk, the hiker walks a distance of km northeast From the components, we get √ √ 5 , b = [5 cos 45◦ , sin 45◦ ] = 2 Thus the net displacement vector is c=a+b= 3+2 + = = 0+3 a + b = √ √ 5 , 4+ 2 a b b a b−c = −2 − (−2) − = = 3 3−3 b c b d − c = −2 − = −2 −5 c 3 4 d d c c 10 a + d = 3 + −2 = 3+3 + (−2) a = −2 d a d 11 2a + 3c = 2[0, 2, 0] + 3[1, −2, 1] = [2 · 0, · 2, · 0] + [3 · 1, · (−2), · 1] = [3, −2, 3] 12 3b − 2c + d = 3[3, 2, 1] − 2[1, −2, 1] + [−1, −1, −2] = [3 · 3, · 2, · 1] + [−2 · 1, −2 · (−2), −2 · 1] + [−1, −1, −2] = [6, 9, −1] CHAPTER VECTORS 13 u = [cos 60◦ , sin 60◦ ] = 2, √ , and v = [cos 210◦ , sin 210◦ ] = − √ √ 3 u+v = − , − , 2 2 √ , − 21 , so that √ √ 3 u−v = + , + 2 2 # » 14 (a) AB = b − a # » # » # » # » (b) Since OC = AB, we have BC = OC − b = (b − a) − b = −a # » (c) AD = −2a # » # » # » (d) CF = −2OC = −2AB = −2(b − a) = 2(a − b) # » # » # » (e) AC = AB + BC = (b − a) + (−a) = b − 2a # » # » # » # » (f ) Note that F A and OB are equal, and that DE = −AB Then # » # » # » # » # » BC + DE + F A = −a − AB + OB = −a − (b − a) + b = 15 2(a − 3b) + 3(2b + a) property e distributivity = (2a − 6b) + (6b + 3a) property b associativity = (2a + 3a) + (−6b + 6b) = 5a 16 property e distributivity −3(a − c) + 2(a + 2b) + 3(c − b) = property b associativity = (−3a + 3c) + (2a + 4b) + (3c − 3b) (−3a + 2a) + (4b − 3b) + (3c + 3c) = −a + b + 6c 17 x − a = 2(x − 2a) = 2x − 4a ⇒ x − 2x = a − 4a ⇒ −x = −3a ⇒ x = 3a 18 x + 2a − b = 3(x + a) − 2(2a − b) = 3x + 3a − 4a + 2b x − 3x = −a − 2a + 2b + b ⇒ ⇒ −2x = −3a + 3b ⇒ 3 x = a − b 2 19 We have 2u + 3v = 2[1, −1] + 3[1, 1] = [2 · + · 1, · (−1) + · 1] = [5, 1] Plots of all three vectors are w v v v u v u 1.1 THE GEOMETRY AND ALGEBRA OF VECTORS 20 We have −u − 2v = −[−2, 1] − 2[2, −2] = [−(−2) − · 2, −1 − · (−2)] = [−2, 3] Plots of all three vectors are u v u w v u 1 v 21 From the diagram, we see that w = −2u + 4v u u w v v v v 1 u 22 From the diagram, we see that w = 2u + 3v u w u v u v v 1 23 Property (d) states that u + (−u) = The first diagram below shows u along with −u Then, as the diagonal of the parallelogram, the resultant vector is Property (e) states that c(u + v) = cu + cv The second figure illustrates this CHAPTER VECTORS cv cu v cu u v cu u u v u cv u v 24 Let u = [u1 , u2 , , un ] and v = [v1 , v2 , , ], and let c and d be scalars in R Property (d): u + (−u) = [u1 , u2 , , un ] + (−1[u1 , u2 , , un ]) = [u1 , u2 , , un ] + [−u1 , −u2 , , −un ] = [u1 − u1 , u2 − u2 , , un − un ] = [0, 0, , 0] = Property (e): c(u + v) = c ([u1 , u2 , , un ] + [v1 , v2 , , ]) = c ([u1 + v1 , u2 + v2 , , un + ]) = [c(u1 + v1 ), c(u2 + v2 ), , c(un + )] = [cu1 + cv1 , cu2 + cv2 , , cuthe line parallel to the given line through P = (−1, 0, 3) is       −1 −1 x y  =  0 + t  3 z −1 24 Since the normal form is n · x = n · p, we use the given information to determine n and p Note that a plane is parallel to a given plane if their normal vectors  are  equal Since the general form of the given plane is 6x − y + 2z = 3, its normal vector is n = −1, so this must be a normal vector of the desired plane as well Furthermore, since our plane passes through the point P = (0, −2, 5), we have   p = −2 So the normal form of the plane parallel to 6x − y + 2z = through (0, −2, 5) is             x 6 x −1 · y  = −1 · −2 or −1 · y  = 12 z z 25 Using Figure 1.34 in Section 1.2 for reference, we will find a normal vector n and a point vector p for each of the sides, then substitute into n · x = n · p to get an equation for each plane (a) Start with P determined by the face of the cube in the xy-plane Clearly a normal vector for P1 is n = 0, or any vector parallel to the x-axis Also, the plane passes through P = (0, 0, 0), 0  so we set p = 0 Then substituting gives         x 0 · y  = 0 · 0 or x = 0 z 0 So the general equation for P1 is x = Applying the same argument above to the plane P2 determined by the face in the xz-plane gives a general equation of y = 0, and similarly the plane P3 determined by the face in the xy-plane gives a general equation of z = Now consider P4 , the plane containing the face parallel to the face  in  the yz-plane but passing through (1, 1, 1) Since P4 is parallel to P1 , its normal vector is also 0; since it passes through   (1, 1, 1), we set p = 1 Then substituting gives         x 1 0 · y  = 0 · 1 or x = z So the general equation for P4 is x = Similarly, the general equations for P5 and P6 are y = and z = ... office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan Locate your local office at: www.cengage.com/global Cengage Learning products are represented... 6.4 Linear Transformations 6.5 The Kernel and Range of a Linear Transformation 6.6 The Matrix of a Linear Transformation Exploration: Tiles, Lattices, and the... that F A and OB are equal, and that DE = −AB Then # » # » # » # » # » BC + DE + F A = ? ?a − AB + OB = ? ?a − (b − a) + b = 15 2 (a − 3b) + 3(2b + a) property e distributivity = ( 2a − 6b) + (6b + 3a)