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NOTE UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD LICENSED AUCTIONED. OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN. READ IMPORTANT LICENSE INFORMATION, Dear Professor or Other Supplement Recipient may not copy or distribute any portion of the Supplement to any. third party You may not sell license auction or otherwise. Cengage Learning has provided you with this product the redistribute the Supplement in any form We ask that you take. Supplement for your review and to the extent that you adopt reasonable steps to protect the Supplement from unauthorized. the associated textbook for use in connection with your course use reproduction or distribution Your use of the Supplement. the Course you and your students who purchase the indicates your acceptance of the conditions set forth in this. textbook may use the Supplement as described below Cengage Agreement If you do not accept these conditions you must. Learning has established these use limitations in response to return the Supplement unused within 30 days of receipt. concerns raised by authors professors and other users. regarding the pedagogical problems stemming from unlimited All rights including without limitation copyrights patents and. distribution of Supplements trade secrets in the Supplement are and will remain the sole and. exclusive property of Cengage Learning and or its licensors The. Cengage Learning hereby grants you a nontransferable license Supplement is furnished by Cengage Learning on an as is basis. to use the Supplement in connection with the Course subject to without any warranties express or implied This Agreement will. the following conditions The Supplement is for your personal be governed by and construed pursuant to the laws of the State. noncommercial use only and may not be reproduced posted of New York without regard to such State s conflict of law rules. electronically or distributed except that portions of the. Supplement may be provided to your students IN PRINT FORM Thank you for your assistance in helping to safeguard the integrity. ONLY in connection with your instruction of the Course so long of the content contained in this Supplement We trust you find the. as such students are advised that they Supplement a useful teaching tool. Printed in the United States of America, 1 2 3 4 5 6 7 17 16 15 14 13. 1 Vectors 3, 1 1 The Geometry and Algebra of Vectors 3. 1 2 Length and Angle The Dot Product 10, Exploration Vectors and Geometry 25. 1 3 Lines and Planes 27, Exploration The Cross Product 41.

1 4 Applications 44, Chapter Review 48, 2 Systems of Linear Equations 53. 2 1 Introduction to Systems of Linear Equations 53. 2 2 Direct Methods for Solving Linear Systems 58, Exploration Lies My Computer Told Me 75. Exploration Partial Pivoting 75, Exploration An Introduction to the Analysis of Algorithms 77. 2 3 Spanning Sets and Linear Independence 79, 2 4 Applications 93. 2 5 Iterative Methods for Solving Linear Systems 112. Chapter Review 123, 3 Matrices 129, 3 1 Matrix Operations 129.

3 2 Matrix Algebra 138, 3 3 The Inverse of a Matrix 150. 3 4 The LU Factorization 164, 3 5 Subspaces Basis Dimension and Rank 176. 3 6 Introduction to Linear Transformations 192, 3 7 Applications 209. Chapter Review 230, 4 Eigenvalues and Eigenvectors 235. 4 1 Introduction to Eigenvalues and Eigenvectors 235. 4 2 Determinants 250, Exploration Geometric Applications of Determinants 263.

4 3 Eigenvalues and Eigenvectors of n n Matrices 270. 4 4 Similarity and Diagonalization 291, 4 5 Iterative Methods for Computing Eigenvalues 308. 4 6 Applications and the Perron Frobenius Theorem 326. Chapter Review 365, 2 CONTENTS, 5 Orthogonality 371. 5 1 Orthogonality in Rn 371, 5 2 Orthogonal Complements and Orthogonal Projections 379. 5 3 The Gram Schmidt Process and the QR Factorization 388. Exploration The Modified QR Process 398, Exploration Approximating Eigenvalues with the QR Algorithm 402. 5 4 Orthogonal Diagonalization of Symmetric Matrices 405. 5 5 Applications 417, Chapter Review 442, 6 Vector Spaces 451.

6 1 Vector Spaces and Subspaces 451, 6 2 Linear Independence Basis and Dimension 463. Exploration Magic Squares 477, 6 3 Change of Basis 480. 6 4 Linear Transformations 491, 6 5 The Kernel and Range of a Linear Transformation 498. 6 6 The Matrix of a Linear Transformation 507, Exploration Tiles Lattices and the Crystallographic Restriction 525. 6 7 Applications 527, Chapter Review 531, 7 Distance and Approximation 537.

7 1 Inner Product Spaces 537, Exploration Vectors and Matrices with Complex Entries 546. Exploration Geometric Inequalities and Optimization Problems 553. 7 2 Norms and Distance Functions 556, 7 3 Least Squares Approximation 568. 7 4 The Singular Value Decomposition 590, 7 5 Applications 614. Chapter Review 625, 8 Codes 633, 8 1 Code Vectors 633. 8 2 Error Correcting Codes 637, 8 3 Dual Codes 641.

8 4 Linear Codes 647, 8 5 The Minimum Distance of a Code 650. 1 1 The Geometry and Algebra of Vectors, H 2 3L 3 H2 3L. 2 3 5 2 2 4 2 2 0 2 3 5, 3 0 3 3 3 0 3 3 0 3 2 5, plotting those vectors gives. 4 CHAPTER 1 VECTORS, 4 Since the heads are all at 3 2 1 the tails are at. 3 0 3 3 3 0 3 1 2 3 1 4, 2 2 0 2 2 0 2 2 4 2 1 3, 1 0 1 1 1 0 1 1 0 1 2 3.

5 The four vectors AB are, In standard position the vectors are. a AB 4 1 2 1 3 3, b AB 2 0 1 2 2 1, c AB 12 2 3 32 32 32. d AB 16 13 12 13 16 16, 1 1 THE GEOMETRY AND ALGEBRA OF VECTORS 5. 6 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 4 km north so a 0 4 During the second part of the. walk the hiker walks a distance of 5 km northeast From the components we get. b 5 cos 45 5 sin 45, Thus the net displacement vector is. 3 2 3 2 5 3, 2 2 2 2 4 3, 9 d c 1 2 3 4 5, 10 a d 1 2 3 4 5 6.

11 2a 3c 2 0 2 0 3 1 2 1 2 0 2 2 2 0 3 1 3 2 3 1 3 2 3. 3b 2c d 3 3 2 1 2 1 2 1 1 1 2, 3 3 3 2 3 1 2 1 2 2 2 1 1 1 2. 6 CHAPTER 1 VECTORS, 13 u cos 60 sin 60 1, and v cos 210 sin 210 23 21 so that. 1 3 3 1 1 3 3 1, 2 2 2 2 2 2 2 2, 14 a AB b a, b Since OC AB we have BC OC b b a b a. 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 0, property e property b.

distributivity associativity, 15 2 a 3b 3 2b a 2a 6b 6b 3a 2a 3a 6b 6b 5a. property e, distributivity, 3 a c 2 a 2b 3 c b 3a 3c 2a 4b 3c 3b. property b, associativity, 3a 2a 4b 3b 3c 3c, 17 x a 2 x 2a 2x 4a x 2x a 4a x 3a x 3a. x 2a b 3 x a 2 2a b 3x 3a 4a 2b, x 3x a 2a 2b b, 19 We have 2u 3v 2 1 1 3 1 1 2 1 3 1 2 1 3 1 5 1 Plots of all three vectors are. 1 1 2 3 4 5 6, 1 1 THE GEOMETRY AND ALGEBRA OF VECTORS 7.

20 We have u 2v 2 1 2 2 2 2 2 2 1 2 2 2 3 Plots of all three. vectors are, 21 From the diagram we see that w 6, 1 1 2 3 4 5 6. 22 From the diagram we see that w 2u 9, 2 1 1 2 3 4 5 6 7. 23 Property d states that u u 0 The first diagram below shows u along with u Then as the. diagonal of the parallelogram the resultant vector is 0. Property e states that c u v cu cv The second figure illustrates this. 8 CHAPTER 1 VECTORS, 24 Let u u1 u2 un and v v1 v2 vn 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, Property e, c u v c u1 u2 un v1 v2 vn.

c u1 v1 u2 v2 un vn, c u1 v1 c u2 v2 c un vn, cu1 cv1 cu2 cv2 cun cvn. cu1 cu2 cun cv1 cv2 cvn, c u1 u2 un c v1 v2 vn, Property f. c d u c d u1 u2 un, c d u1 c d u2 c d un, cu1 du1 cu2 du2 cun dun. cu1 cu2 cun du1 du2 dun, c u1 u2 un d u1 u2 un, Property g. c du c d u1 u2 un, c du1 du2 dun, cdu1 cdu2 cdun, cd u1 cd u2 cd un.

cd u1 u2 un, 25 u v 0 1 1 1 1 0, 26 u v 1 1 0 1 1 1 0 0 1. 1 1 THE GEOMETRY AND ALGEBRA OF VECTORS 9, 27 u v 1 0 1 1 1 1 1 1 0 1 0 0. 28 u v 1 1 0 1 0 0 1 1 1 0 1 0 1 0 0, 0 1 2 3 0 1 2 3. 0 0 1 2 3 0 0 0 0 0, 1 1 2 3 0 1 0 1 2 3, 2 2 3 0 1 2 0 2 0 2. 3 3 0 1 2 3 0 3 2 1, 0 1 2 3 4 0 1 2 3 4, 0 0 1 2 3 4 0 0 0 0 0 0.

1 1 2 3 4 0 1 0 1 2 3 4, 2 2 3 4 0 1 2 0 2 4 1 3, 3 3 4 0 1 2 3 0 3 1 4 2. 4 4 0 1 2 3 4 0 4 3 2 1, 31 2 2 2 6 0 in Z3, 32 2 2 2 3 2 0 in Z3. 33 2 2 1 2 2 2 3 1 1 1 in Z3, 34 3 1 2 3 4 2 1 1 in Z4. 35 2 3 2 4 3 0 0 in Z4, 36 3 3 3 2 4 6 0 0 in Z4, 37 2 1 2 2 1 2 in Z3 2 1 2 2 1 0 in Z4 2 1 2 2 1 3 in Z5. 38 3 4 3 2 4 2 2 1 2 in Z5, 39 8 6 4 3 8 4 5 in Z9.

40 2100 210 1024 10 110 1 in Z11, 41 2 1 2 2 0 1 1 1 0 in Z33. 42 2 2 2 1 2 2 2 2 2 1 1 1 2 in Z33, 43 2 3 1 1 2 3 3 2 1 2 2 0 3 3 2 2 2 0 2 3 2 3 0 0 2 2 in Z44. 2 3 1 1 2 3 3 2 1 2 1 4 3 3 2 1 2 4 2 3 2 3 2 3 1 1 in Z45. 44 x 2 3 2 2 4 in Z5, 45 x 1 5 1 1 2 in Z6, 46 x 2 1 2 in Z3. 47 No solution 2 times anything is always even so cannot leave a remainder of 1 when divided by 4. 48 x 2 1 3 in Z5, 49 x 3 1 4 2 4 3 in Z5, 50 No solution 3 times anything is always a multiple of 3 so it cannot leave a remainder of 4 when. divided by 6 which is also a multiple of 3, 51 No solution 6 times anything is always even so it cannot leave an odd number as a remainder when.

divided by 8, 10 CHAPTER 1 VECTORS, 52 x 8 1 9 7 9 8 in Z11. 53 x 2 1 2 3 3 2 2 2 in Z5, 54 No solution This equation is the same as 4x 2 5 3 3 in Z6 But 4 times anything is even. so it cannot leave a remainder of 3 when divided by 6 which is also even. 55 Add 5 to both sides to get 6x 6 so that x 1 or x 5 since 6 1 6 and 6 5 30 6 in Z8. 56 a All values b All values c All values, 57 a All a 6 0 in Z5 have a solution because 5 is a prime number. b a 1 and a 5 because they have no common factors with 6 other than 1. c a and m can have no common factors other than 1 that is the greatest common divisor gcd of. a and m is 1, 1 2 Length and Angle The Dot Product. 1 Following Example 1 15 u v 1 3 2 1 3 2 1, 2 Following Example 1 15 u v 3 4 2 6 12 12 0.

3 u v 2 3 1 2 2 3 3 1 2 6 3 11, 4 u v 3 2 1 5 0 6 4 1 1 4 0 2 4 8 2 46 0 28 2 62. 3 0 1 4 2 2 3 0 0 5 4 2 2, 2 07 4 33 1 12 2 29 3 25 1 72 2 07 4 33 1 83 1 54 3 6265. 7 Finding a unit vector v in the same direction as a given vector u is called normalizing the vector u. Proceed as in Example 1 19 p, kuk 1 2 22 5, so a unit vector v in the same direction as u is. 8 Proceed as in Example 1 19, kuk 32 2 2 9 4 13, so a unit vector v in the direction of u is. kuk 13 2 213, 1 2 LENGTH AND ANGLE THE DOT PRODUCT 11.

9 Proceed as in Example 1 19 p, kuk 12 22 32 14, so a unit vector v in the direction of u is. 10 Proceed as in Example 1 19, kuk 3 22 0 6 2 1 4 2 10 24 0 36 1 96 12 56 3 544. so a unit vector v in the direction of u is, v u 0 4 0 169. 11 Proceed as in Example 1 19, kuk 12 2 3 02 6, so a unit vector v in the direction of u is. 12 Proceed as in Example 1 19, kuk 1 122 3 25 2 2 072 1 83 2 1 2544 10 5625 4 2849 3 3489.

19 4507 4 410, so a unit vector v in the direction of u is. v u 1 12 3 25 2 07 1 83 0 254 0 737 0 469 0 415, 13 Following Example 1 20 we compute u v so. d u v ku vk 4 12 17, 14 Following Example 1 20 we compute u v so. d u v ku vk 1 8 65, 15 Following Example 1 20 we compute u v 2 3 1 so. d u v ku vk 1 1 22 6, 12 CHAPTER 1 VECTORS, 3 2 1 5 1 7.

16 Following Example 1 20 we compute u v 0 6 4 1 4 7 so. 1 4 0 2 1 2, d u v ku vk 1 72 4 7 2 1 2 2 26 42 5 14. 17 a u v is a real number so ku vk is the norm of a number which is not defined. b u v is a scalar while w is a vector Thus u v w adds a scalar to a vector which is not a. defined operation, c u is a vector while v w is a scalar Thus u v w is the dot product of a vector and a scalar. which is not defined, d c u v is the dot product of a scalar and a vector which is not defined. 18 Let be the angle between u and v Then, u v 3 1 0 1 3 2. note under no circumstances may this material or any portion thereof be sold licensed auctioned or otherwise redistributed except as may be permitted by the license terms herein