Elementary Linear Algebra 6e - Larson, Edwards, Falvo - Answer Key

Elementary Linear Algebra 6e - Larson, Edwards, Falvo - Answer Key

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Unformatted text preview: Answer Key A9 ANSWER KEY Chapter 1 Section 1.1 (page 11) 1. Linear 7. x 2t y t 11. x1 x2 17. 4 3 2 1 -4 -2 -2 -4 25. 6 4 -2 -4 -6 -8 -10 -12 y 27. 2x - y = 12 x 1 2 3 6 7 8 6 4 2 y 0.05x - 0.03y = 0.07 x 3. Not linear 9. x y z 13. x y z y 3 2 3 2 5. Not linear 1 s t s t x y 29. x+3 y-1 + =1 4 3 -2 -4 -6 1 2 4 5 6 7 8 0.07x + 0.02y = 0.16 5 2 y 8 6 4 2 x 1 3 4 5 3 0 19. 15. x1 x2 x3 y 4 3 1 t 2t t x y x y 18 5 3 5 2 1 - 2x + 2y = 5 x y + =1 4 6 x-y=2 x 3 4 x-y=1 x 1 2 3 4 -1 -2 -4 -2 -3 -4 x-y=3 2x + y = 4 31. (a) -3x - y = 3 (b) Inconsistent 3 x y 21. 10 8 6 4 2 -2 2 0 y No solution -4 4 23. 2x + y = 9 -6 - 4 - 2 -2 -4 y x 4 6 -3 5x - y = 11 6x + 2y = 1 2x - 8y = 3 2x - y = 5 3x - 5y = 7 x 2 4 6 8 10 - 12 33. (a) 3 (b) Consistent (c) x 1 2 y 1 4 1 2 1 4 -4 4 (d) x y x y 4 1 x y 2 1 -3 1 2x + y = 0 (e) The solutions are the same. A10 Answer Key 4x - 8y = 9 35. (a) 3 -4 4 (b) Consistent (c) There are infinite solutions. (d) x 9 2t 4 y t (e) The solutions are consistent. 0.8x - 1.6y = 1.8 -3 37. x1 x2 43. x y 1 1 7 1 39. u v 45. x1 x2 51. x1 x2 x3 57. x1 x2 x3 x4 40 40 8 7 5 2 1 2t 41. x y 47. x y z 1 59. x y z 1 3 2 3 1 2 3 49. No solution 53. No solution 4t t 15 40 45 75 55. x y z w 61. x1 x2 x3 63. x y z w 1 0 3 2 1 5 4 5 1 2 1.2 0.6 2.4 69. (a) True. You can describe the entire solution set using parametric representation. ax by c Choosing y t as the free variable, the solution is b c t, y t, where t is any real number. x a a (b) False. For example, consider the system x1 x 2 x3 1 x1 x2 x3 2 which is an inconsistent system. (c) False. A consistent system may have only one solution. 4 71. 3x1 x2 3x1 x2 4 (The answer is not unique.) 73. x 3 y 4 75. x y 2 5 1 t 6.8813 163.3111 210.2915 59.2913 65. This system must have at least one solution because x y z 0 is an obvious solution. Solution: x 0 y 0 z 0 This system has exactly one solution. 67. This system must have at least one solution because x y z 0 is an obvious solution. 3 Solution: x 5t 4 y 5t z t This system has an infinite number of solutions. 4t 1 1 1 z where t 5, , 0 t 4 2 79. k 77. x cos y sin 8 81. All k 1 83. k 3 85. k 1, 2 87. (a) Three lines intersecting at one point (b) Three coincident lines (c) Three lines having no common point 89. Answers will vary. (Hint: Choose three different values of x and solve the resulting system of linear equations in the variables a, b, and c.) Answer Key A11 91. x 5x 4y 6y y 5 4 3 2 3 13 x 4y 14y y 5 3 28 27. x y 33. x y z w 35. x y z 39. x1 x2 x3 x4 x5 x 4 2 x - 4y = - 3 x 3 4 5 14y = 28 4 3 x - 4 -2 -4 -5 5x - 6y = 13 - 4 -2 -2 -3 -4 -5 x - 4y = - 3 3 4 5 x 4y y y 5 4 3 3 2 x y 5 2 y 5 4 3 1 x=5 y=2 y=2 x 3 4 5 - 4 -2 -2 -3 -4 -5 x - 4y = -3 -4 -2 -2 -3 -4 -5 1 2 3 4 43. x1 x2 x3 47. (a) (b) (c) (d) The intersection points are all the same. 93. x 39,600 y 398 The graphs are misleading because, while they appear parallel, when the equations are solved for y, they have slightly different slopes. 37. x1 23.5361 0.5278t x2 18.5444 4.1111t x3 7.4306 2.1389t x4 t 41. x1 2 1 x2 2 2 x3 3 6 5 3 x4 x5 1 4 x6 3 45. x1 0 t x2 s t x3 0 t x4 t Two equations in two variables 4 All real k 3 Two equations in three variables All real k 100 s 54 52t t 0 2 4t t 29. x1 x2 x3 96t 3s 4 3 2 31. x1 x2 x3 1 2 t 2t 3t Section 1.2 (page 26) 1. 9. 11. 13. 15. 3 3 3. 2 4 5. 1 Reduced row-echelon form Not in row-echelon form Not in row-echelon form x1 0 2 17. x1 x2 2 x2 1 x3 1 26 13 7 4 23. x y 3 2 5 7. 4 5 49. (a) a b c 0 (b) a b c 0 (c) Not possible 51. (a) x y z (c) x y z 1 53. 0 1 55. 0 8 3 8 3 5 6t 5 6t t 3 3 (b) x y z 18 7 11 14 t 20 13 7 14 t t t t 19. x1 x2 x3 1 1 0 21. x1 x2 x3 x4 25. No solution t 0 1 0 1 , 1 0 (d) Each system has an infinite number of solutions. k 0 , 0 0 1 0 , 0 0 0 0 A12 Answer Key 57. (a) True. In the notation m n, m is the number of rows of the matrix. So, a 6 3 matrix has six rows. (b) True. At the top of page 19, the sentence reads, "It can be shown that every matrix is row-equivalent to a matrix in row-echelon form." (c) False. Consider the row-echelon form 1 0 0 0 0 0 1 0 0 1 0 0 1 0 2 0 0 0 1 3 which gives the solution x1 0, x2 1, x3 2, and x4 3. (d) True. Theorem 1.1 states that if a homogeneous system has fewer equations than variables, then it must have an infinite number of solutions. 59. ad bc 0 1, 3 61. 63. Yes, it is possible: x1 x2 x3 0 x1 x2 x3 1 65. The rows have been interchanged. The first elementary row operation is redundant, so you can just use the second and third elementary row operations. 67. An inconsistent matrix in row-echelon form would have a row consisting of all zeros except for the last entry. 69. In the matrix in reduced row-echelon form, there would be zeros above any leading ones. 3. (a) p x (b) y 10 8 6 4 2 2x (5, 10) (3, 6) (2, 4) x 1 2 3 4 5 5. (a) p z px (b) 7 7 y 12 9 7 2z 7 2 x 3 2 2z 2007 3 2 x 2007 2 (1, 12) (0, 7) (-1, 5) -1 3 z 1 (2006) (2007) (2008) 7. y is not a function of x because the x-value of 3 is repeated. 9. p x 11. p x 1 x 3x x3 y= 1 1+x y y (- 1, 2) 5 4 3 x -1 Section 1.3 (page 38) 1. (a) p x (b) y 29 18x 3x2 (0, 1) -1 y= x2 15 - 7x 15 +1 -1 -2 1 2 )2, 1) )4, 1 ) 3 5 x 1 2 3 4 (1, -2) 6 5 (2, 5) 4 3 2 1 (4, 5) 249 2.7z 0.05z2 where z 13. p z Year 2010: p 323 million Year 2020: p 375 million 15. (a) a0 a0 a0 a0 a1 3a1 5a1 7a1 a2 9a2 25a2 49a2 a3 27a3 125a3 343a3 x 1990 (3, 2) x 1 2 3 4 5 6 10,003 10,526 12,715 14,410 Answer Key A13 (b) p z 11,041.25 1606.5z 613.25z2 45z3 where z x 2000 No, the profits have increased every year except 2006 and our model predicts a decrease in 2008. This is not a reasonable estimate. 4 2 4 x x 17. p x 2 8 0.889 3 9 (Actual value is 3 2 sin 19. Solve the system: p 1 a0 a1 p0 a0 a0 a1 p1 a0 a1 21. (a) x1 s x2 t x3 600 s x4 s t x5 500 t x6 s x7 t 23. (a) x1 x2 x3 x4 25. I1 I2 I3 29. 31. 1 x 1 2 2 2 4 1 0 1 1 a2 a2 a2 35. x2 x3 13 x2 3x3 46 x2 x3 0 where x1 touchdowns, x2 and x3 field goals x1 5 x2 4 x3 4 x1 6x1 extra points, Review Exercises Chapter 1 (page 41) 0.866.) 0 0 0 0 0 0 600 0 500 0 0 (c) x1 x2 x3 x4 x5 x6 x7 0 500 600 500 1000 0 500 200 0 300 100 1. Not linear 5. Not linear 1 9. x 4 y s z t 1 11. x 2 y 19. x1 x2 3 2 3. Linear 7. Linear 1 2s 3 2t (b) x1 x2 x3 x4 x5 x6 x7 13. x y 21. x y 0 0 12 8 15. x y 23. 2 0 0 3 17. No solution 1 2 4 5 100 t (b) x1 100 t x2 200 t x3 t x4 27. (a) I1 I2 I3 3 2 x 1 x 1 2 4 x 2 x 2 1 2 1 2 100 (c) x1 100 x2 x3 200 0 x4 (b) I1 0 I2 1 I3 1 25. Row-echelon form (not reduced) 27. Not in row-echelon form 29. x1 x2 x3 35. x y z 41. x y z 47. x1 x2 x3 2t t 0 4 5 t 0 2 t 0 0 0 3t 2t 31. x y z 37. x y z 3 2 2 3 3 1 t 2t 2t 33. x y z 39. x1 x2 x3 x4 45. x y z w 51. k 1 2 1 3 1 1 4 3 2 2t t t 0 1 x 33. x y 2 4t 43. x 1 y 0 z 4 w 2 4t 49. x1 1 x2 2t x3 t A14 Answer Key 3 53. (a) b 2a and a (b) b 2a 3 and b 6 (c) a 55. Use an elimination method to get both matrices in reduced row-echelon form. The two matrices are row-equivalent because each is row-equivalent to 1 0 0 0 1 0 . 0 0 1 1 0 1 2 . . . 2 n 0 1 2 3 . . . n 1 0 0 . . . 0 57. 0 0 . . . . . . . . . 0 0 0 0 0 59. (a) False. See page 3, following Example 2. (b) True. See page 5, Example 4(b). 61. (a) 3x1 2x2 x3 59 3x1 x2 0 x2 x3 1 where x1 number of three-point baskets, x2 number of two-point baskets, x3 number of one-point free throws (b) x1 5 x2 15 x3 14 63. 1 x 2 x 90 2 2 x 135 2 x 69. (a) a0 a0 a0 4a1 80a1 16a2 6400a2 80 25 8 1 32 25 8 x 80 68 30 (b) and (c) a0 a1 a2 1 80. So, y 32 x2 (d) The results to (b) and (c) are the same. (e) There is precisely one polynomial function of degree n 1 (or less) that fits n distinct points. 71. I1 I2 I3 5 13 6 13 1 13 Chapter 2 Section 2.1 (page 56) 1. (a) 3 1 0 5 7 1 2 11 5 7 4 8 2 4 8 2 2 7 1 10 3 9 15 6 3 0 0 7 2 2 10 4 (b) 1 3 5 2 0 9 3 2 15 2 (c) 2 4 2 2 (d) (e) 0 5 3 4 4 0 1 2 1 2 2 3. (a) (b) 5 1 5 7 2 (c) 12 4 6 2 8 10 65. (a) p x y (b) 25 20 15 10 5 25 2 2x (4, 20) (d) (e) 7 25 2 (2, 5) (3, 0) x 1 2 3 4 5 3 5. (a) 7 2 6 (c) 4 0 (b) 3 3 2 6 1 2 0 0 0 2 4 1 2 3 2 3 8 4 15 5 2 50 67. p x 2x 2x (First year is represented by x 95 Fourth-year sales: p 3 0.) (d) Answer Key A15 3 2 3 6 3 2 (e) 6 2 7. (a) c21 9. x 11. (a) 3, y 0 6 1 2 9 2 1 6 2, z (b) c13 1 (b) 2 31 2 14 4 16 46 0 (d) 29 21. 3 4 1 2 8 15 4 16 20 10 16 19 0 6 9 26 23. 4 21 20 12 12 1 3 2 4 0 4 8 4 36 7 6 21 6 2 3 26 33 8 6 5 11 6 9 2 28 8 11 6 23 33 25. 3 15 12 27. Not defined, sizes do not match. 29. 1 1 x1 x2 x1 x2 3 x1 1 x2 x1 x2 2 3 5 3 1 5 13. (a) Not defined 1 4 0 60 20 10 60 5 0 6 5 12 5 1 7 3 1 1 1 2 6 1 4 8 2 2 5 8 0 1 2 8 13 3 7 4 8 2 27 22 4 4 11 19 27 14 72 24 12 72 3 (b) 10 26 15. (a) (b) Not defined 31. 2 6 17. (a) (b) Not defined 33. 8 8 9 3 8 5 3 3 1 5 3 1 8 1 7 1 17 9 1 1 2 x1 x2 x3 x1 x2 x3 9 6 17 1 1 2 20 8 16 1 3 2 19. (a) 5 5 1 9 7 10 13 1 10 6 7 9 5 10 34 11 11 30 6 1 4 2 8 10 11 11 2 6 4 2 2 2 15 1 9 10 1 3 7 3 6 6 3 2 0 2 11 11 3 21 37 2 10 3 35. 1 3 0 5 1 2 2 1 5 x1 x2 x3 x1 x2 x3 (b) 37. 39. a 41. w 1 43. 0 0 5 3 7, b z, x 0 4 0 2 1 4, c y 0 0 9 45. AB BA 10 0 10 0 0 12 0 12 1 2, d 7 2 (c) A16 Answer Key 47. Proof 55. Let A 49. 2 51. 4 53. Proof a11 a12 . a21 a22 Then the given matrix equation expands to a11 a21 a12 a22 1 0 . a11 a21 a12 a22 0 1 Because a11 a21 1 and a11 a21 0 cannot both be true, you can conclude that there is no solution. i2 0 1 0 57. (a) A2 0 i2 0 1 3 0 i i 0 A3 0 i3 0 i A4 (b) B2 i4 0 i2 0 0 i4 0 i2 1 0 0 1 1 0 0 1 4 0 71. 1 0 0 5 1 1 2 73. b 3 0 2 3 3 1 (The answer is not unique.) 75. b 1 1 1 2 2 1 0 1 0 5 1 1 1 1 0 1 7 3 1 0 Section 2.2 (page 70) 1. 3 13 2 4 3 4 3 10 3 3. 2 3 11 3 0 12 12 24 5. 13 3 7 28 10 3 7 14 I 7. (a) (b) 4 26 3 13 6 1 3 5 16 3 0 4 17 2 10 5 15. AC 19. 0 2 1 3 5 1 4 1 3 18 9 1 0 11. BC 0 1 8 8 0 59. Proof 61. Proof 84 60 30 63. 42 120 84 65. $1037.50 $1400.00 $1012.50 Each entry represents the total profit at each outlet. 67. (a) True. On page 51, ". . . for the product of two matrices to be defined, the number of columns of the first matrix must equal the number of rows of the second matrix." (b) True. On page 55, ". . . the system Ax b is consistent if and only if b can be expressed as . . . a linear combination, where the coefficients of the linear combination are a solution of the system." 69. PP 0.75 0.20 0.30 0.15 0.60 0.40 0.10 0.75 0.20 0.20 0.30 0.30 0.135 0.20 0.20 0.15 0.60 0.40 0.10 0.20 0.30 (c) 14 7 17 5 5 4 4 1 17 6 10 3 (d) 0 1 2 2 2 6 2 3 3 21. 4 0 2 9. 13. 3 2 12 8 1 8 17. Proof 4 2 1 21 3 2 1 3 14 3 1 2 0 2 0 23. (a) 16 (b) 8 4 (c) 0.6225 0.2425 0.33 0.47 0.395 0.405 25. (a) This product represents the changes in party affiliation after two elections. (c) 0 2 1 1 9 5 (b) 5 6 5 6 21 3 5 3 11 Answer Key A17 27. (a) 0 4 3 2 68 26 10 6 29 14 5 5 8 4 0 1 26 41 3 1 14 81 3 2 2 3 5 1 10 3 43 5 5 3 39 13 0 0 3 2 6 1 5 10 5 2 13 13 4755. Proof 59. Symmetric 63. (a) 1 A AT 2 1 2 57. Skew-symmetric 61. Proof a1n a 2n . . . ann a11 a21 . . . a12 a22 . . . . . . . . . a1n a2n . . . an1 an2 . . . ann (b) a11 a12 . . . a21 a22 . . . . . . . . . an1 an2 . . . 1 2 (c) 2a11 a12 a21 . . . a1n an1 . . . a2n an2 a21 a12 2a22 . . . . . . . . . . . . an1 a1n an2 a2n 2ann AT a11 a12 . . . a1n a21 a22 . . . a 2n . . . . . . . . . . . . ann an1 an2 0 a12 a21 . a12 . . an1 a1n an2 AT 1 2 1 2 1 2 (b) 1 2 A 1 2 A2 BA AB B2, which is not 29. A B A B necessarily equal to A2 B2 because AB is not necessarily equal to BA. 2 5 31. AB T BTAT 4 1 33. AB 35. (a) (b) (c) (d) 37. (a) (b) T a11 a21 . . . an1 a12 a22 . . . an2 . . . . . . . . . . . . ann a1n a2n an1 an2 BTAT 4 10 1 0 4 1 4 2 3 1 2 0 . . . a21 . . . a1n . . . a2n a2n . . . AT 1 6 1 2 7 2 1 2 . . . 0 True. See Theorem 2.1, part 1. True. See Theorem 2.3, part 1. False. See Theorem 2.6, part 4, or Example 9. True. See Example 10. a 3 and b 1 a b 1 b 1 a 1 No solution cb 0 c 0a (c) a b c 0; a b c 0 a c 0 3t; let t 1: a 3, b 1, c 1 (d) a b t c t 1 0 0 3 0 1 0 39. 0 41. 0 2 0 0 1 4 0 1 1 43. 45. 1 1 8 2 (c) Proof (d) A 1 A 2 0 4 4 0 1 2 1 2 A 2 1 7 2 0 1 1 0 Skew-symmetric Symmetric 65. (a) A 0 0 1 1 , B 1 0 1 (The answer is not unique.) (b) Proof Section 2.3 (page 84) 1. AB 1 0 1 0 0 0 1 0 1 0 BA 0 0 1 3. AB BA A18 Answer Key 5. 9. 7 3 1 3 3 3 2 9 2 2 1 1 2 3 3 2 7 2 7. 1 1 2 1 3 1 0 0 1 5 19 4 33 7 (c) 37. x 41. 1 16 7 28 30 1 2 1 4 0 40 14 34 14 25 39. x (d) 6 1 8 4 6 1 2 2 4 3 8 2 11. Singular 4 1 3 4 43. Proof 13. 15. 1 1 17. 3 4 7 20 1 0 1 4 1 4 0 10 10 2 4 2 0.8 4.4 3.2 45. (a) (b) (c) (d) 19. Singular 2 1 2 1 (b) x y (b) x1 x2 x3 2 4 0 1 1 31. x1 x2 x3 x4 x5 x6 1 2 3 0 1 2 7 6 1 7 2 5 2 21. 24 10 29 12 1 7 3 7 3 1 0 3 1 23. Singular 25. (a) x y 27. (a) x1 x2 29. x1 x2 x3 x4 x5 x3 0 1 2 1 0 35 4 31 56 1 16 1 1 1 1 (c) x y 3 2 3 4 True. See Theorem 2.7. True. See Theorem 2.10, part 1. False. See Theorem 2.9. True. See "Finding the Inverse of a Matrix by Gauss-Jordan Elimination," part 2, page 76. 47 53. Proof 55. The sum of two invertible matrices is not necessarily invertible. For example, let 1 0 1 0 and B A . 0 1 0 1 0 1 0 2 0 0 1 57. (a) (b) 0 0 0 3 0 3 1 0 0 0 0 4 2 59. (a) Proof (b) H 0 1 0 1 0 0 0 0 1 61. A PDP 1 No, A is not necessarily equal to D. 63. 1 1 0 0 Section 2.4 (page 96) 1. 3. 5. 7. Elementary, multiply Row 2 by 2. Elementary, add 2 times Row 1 to Row 2. Not elementary Elementary, add 5 times Row 2 to Row 3. 0 0 1 0 0 1 1 0 1 0 9. 0 11. 0 1 0 0 1 0 0 33. (a) (c) 17 10 40 1 56 26 34 (b) (d) 84 71 3 2 5 3 (b) 1 4 35. (a) 138 37 24 4 2 3 6 2 8 1 4 2 Answer Key A19 13. 0 1 1 k 0 0 1 0 0 1 1 3 2 0 15. 0 1 1 19. 0 0 1 23. 0 0 0 2 0 1 0 0 1 0 1 0 0 0 1 0 0 0 1 0 0 0 0 2 0 0 0 1 1 0 3 1 0 0 1 0 0 0 1 0 1 6 1 0 0 0 1 0 1 4 1 24 1 4 (c) True. See "Definition of Row Equivalence," page 90. (d) True. See Theorem 2.13. 35. (a) EA will have two rows interchanged. (The same rows are interchanged in E.) (b) E 2 In 37. A 1 17. 1 b 0 a ab 0 1 2 0 1 1 21. 0 0 1 c 1 0 1 1 1 2 1 . 3 1 2 0 39. No. For example, 41. 1 0 1 1 1 1 1 0 1 0 (The answer is not unique.) 1 1 1 0 1 27. 0 1 3 1 0 (The answer is not unique.) 1 0 0 1 2 1 1 0 0 1 29. 0 0 1 0 0 (The answer is not unique.) 25. 1 0 31. 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 2 1 0 1 (The answer is not unique.) 1 0 0 3 0 43. 2 1 0 0 1 1 1 1 0 0 (The answer is not unique.) 45. (a) 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 2 1 0 1 0 0 1 1 2 1 0 0 (The answer is not unique.) 1 2 5 (c) x 1 1 2 0 1 3 1 3 1 3 5 3 (b) y 47. First, factor the matrix A LU. Then, for each righthand side bi , solve Ly bi and Ux y. 49. Idempotent 51. Not idempotent 53. Not idempotent 55. Case 1: b 1, a 0 Case 2: b 0, a any real number 57 61. Proofs (The answer is not unique.) 33. (a) True. See "Remark" following "Definition of Elementary Matrix," page 87. (b) False. Multiplication of a matrix by a scalar is not a single elementary row operation so it cannot be represented by a corresponding elementary matrix. Section 2.5 (page 112) 1. Not stochastic 5. Stochastic 3. Stochastic 7. Next month: 350 people In 2 months: 475 people A20 Answer Key 1 3 9. In 1 month In 2 months Nonsmokers 5025 5047 Smokers of less than 1 pack/day 2500 2499 Smokers of more than 1 pack/day 2475 2454 11. Tomorrow: 25 students 13. Proof In 2 days: 44 students In 30 days: 40 students 15. Uncoded: 19 5 12 , 12 0 3 , 15 14 19 , 15 12 9 , 4 1 20 , 5 4 0 Encoded: 48, 5, 31, 6, 6, 9, 85, 23, 43, 27, 3, 15, 115, 36, 59, 9, 5, 4 17. Uncoded: 3 15 , 13 5 , 0 8 , 15 13 , 5 0 , 19 15 , 15 14 Encoded: 48, 81, 28, 51, 24, 40, 54, 95, 5, 10, 64, 113, 57, 100 19. HAPPY_NEW_YEAR 21. ICEBERG_DEAD_AHEAD 23. MEET_ME_TONIGHT_RON 3 2 2 25. A 1 4 2 3 ; 2 1 1 _SEPTEMBER_THE_ELEVENTH_WE_WILL _ALWAYS_REMEMBER Coal Steel 35. y 39. y 43. (a) (b) 45. (a) (b) 37. y 2x 41. y 0.412 x 3 y 11,650 2400x 3490 gallons y y 3.24t 3.24t 223.5 223.5 1.3 0.6x 0.5x 7.5 Review Exercises Chapter 2 (page 115) 1. 4 5. 0 0 9. x1 2x1 11. 2 3 13 0 6 6 0 x2 3x2 2x2 1 2 3 3 2 1 2 3 14 4 1 1 3 1 1 3 3 x y 1 3 3 x1 x2 x3 0 1 , ATA 2 4 5 1 , ATA 3 9 3 1 3 1 3 20 3 20 1 30 1 5 1 10 2 15 1 5 8 11 3 10 6 18 19 14 3. 14 36 7. 5x x 1 0 2 5 4 10 22 2 1 2 3 2 5 4 2 10 12 4y y 8 40 48 2 22 2x3 x3 4x3 2 13. 2 4 15. AT 27. D 0.1 0.8 0.2 Coal 0.1 Steel Farmer Baker Grocer X 20,000 Coal 40,000 Steel 3 4 , 13 29. X 31. (a) 8622.0 4685.0 3661.4 y 4 3 2 AAT 17. AT 33. (a) 4 y 11 (0, 4) (2, 3) AAT (1, 3) 3 2 (- 2, 0) -1 (0, 1) x 1 4 3 3 4x 1 2 -1 (1, 1) 1 (2, 0) 2 3 19. x 1 2 21. 3 10 1 5 (b) y (c) 1 6 (b) y (c) 2 4 2x Answer Key A21 23. 5 1 4 1 x1 x2 A x1 x2 1 2 22 1 9 1 9 4 9 5 9 47. x 4, y 1, z 1 49. (a) False. See Theorem 2.1, part 1, page 61. (b) True. See Theorem 2.6, part 2, page 68. 51. (a) False. The matrix 1 0 is not invertible. 0 0 (b) False. See Exercise 55, page 72. 10 12 1 2 4 1 3 1 3 1 3 25. 1 2 5 1 3 4 2 15 1 15 7 15 2 x1 1 x2 2 x3 2 5 4 5 3 5 5455 128.2 53. (a) 3551 77.6 7591 178.6 The first column of the matrix gives the total sales for each type of gas and the second column gives the profit for each type of gas. (b) $384.40 2 2 3 473.5 588.5 55. (a) B (b) BA (c) The matrix BA represents the numbers of calories burned by the 120-pound person and the 150-pound person. 1 A 1 x1 x2 x3 27. 1 31. 0 0 1 14 1 21 1 42 2 21 2 3 3 29. x 4 0 1 3 0 1 0 1 3 2 0 0 1 0 1 (The answer is not unique.) 33. 57. Not stochastic 80 68 65 , P 2X , P 3X 59. PX 112 124 127 110,000 Region 1 123,125 Region 1 61. (a) 100,000 Region 2 (b) 100,000 Region 2 90,000 Region 3 76,875 Region 3 63. Uncoded: 15 14 , 5 0 , 9 6 , 0 2 , 25 0 , 12 1 , 14 4 Encoded: 103, 44, 25, 10, 57, 24, 4, 2, 125, 50, 62, 25, 78, 32 3 2 ; ALL_SYSTEMS_GO 65. A 1 4 3 2 1 0 0 1 1 ; INVASION_AT_DAWN 67. A 1 5 3 3 69. _CAN_YOU_HEAR_ME_NOW 71. D 73. y 20 3 1 0 0 1 0 0 1 0 1 1 0 0 1 2 0 1 0 35. 0 0 0 4 0 0 1 0 0 1 (The answer is not unique.) 1 0 1 0 37. and 0 1 0 1 (The answer is not unique.) 0 0 1 0 1 0 , , and 39. 0 0 0 1 0 0 (The answer is not unique.) 1 41. (a) a (b) Proof b 1 c 1 1 0 2 5 43. Proof 45. 3 1 0 1 (The answer is not unique.) 0.20 0.30 3 2x 0.50 , X 0.10 133,333 133,333 75. y 2 5 9 5x A22 Answer Key 77. (a) y 19 14x (b) 41.4 kilograms per square kilometer 79. (a) 1.828x (c) x 0 1 2 3 4 30.81 Actual 30.37 32.87 34.71 36.59 38.14 (b) 1.828x 30.81 The models are the same. Estimated 30.81 32.64 34.47 36.29 38.12 (b) C11 C21 C31 23 5 7 C12 C22 C32 8 5 22 C13 C23 C33 22 5 23 5 39.63 39.95 The estimated values are very close to the actual values. (d) 49.09 (e) 2011 81. (a) 0.12x 1.9 (b) 0.12x 1.9 The models are the same. (c) x 0 1 2 Actual 1.8 2.1 2.3 Estimated 1.9 2.0 2.1 3 2.4 2.3 4 2.3 2.4 5 2.5 2.5 The estimated values are close to the actual values. (d) 3.1 million (e) 2015 17. (a) 4 5 5 5 6 5 75 (b) 2 8 5 5 3 22 75 19. 58 21. 30 23. 0.022 25. 4x 2y 2 27. 168 29. 0 31. 65,644w 62,256x 12,294y 24,672z 33. 100 35. 43.5 37. 1098 39. 329 41. 24 43. 0 45. 30 47. (a) False. See "Definition of the Determinant of a 2 2 Matrix," page 123. (b) True. See the first line after "Remark," page 124. (c) False. See "Definitions of Minors and Cofactors of a Matrix," page 124. 49. x 51. x 0, 1 1, 4 53. x 55. 1, 4 1 3 57. 59. 8uv 1 2, 0, or 1 61. e5x 63. 1 ln x 65. Expanding along the first row, the determinant of a 4 4 matrix involves four 3 3 determinants. Each of these 3 3 determinants requires six triple products. So, there are 4 6 24 quadruple products. 67. wz xy 69. wz xy 71. xy2 xz2 yz2 x2y x2z y2z 73. bc2 ca2 ab2 ba2 ac2 cb2 75. (a) Proof x 1 0 0 0 x 1 0 0 0 x 1 d c b a Chapter 3 Section 3.1 1. 1 9. 6 13. (a) M11 M12 M21 M22 15. (a) M11 M21 M31 (page 130) (b) 3. 5 4 3 2 1 23 5 7 M12 M22 M32 5. 27 11. 2 (b) C11 C12 C21 C22 8 5 22 7. 4 4 3 2 1 M13 M23 M33 22 5 23 5 24 Section 3.2 (page 140) 1. The first row is 2 times the second row. If one row of a matrix is a multiple of another row, then the determinant of the matrix is zero. 3. The second row consists entirely of zeros. If one row of a matrix consists entirely of zeros, then the determinant of the matrix is zero. Answer Key A23 5. The second and third columns are interchanged. If two columns of a matrix are interchanged, then the determinant of the matrix changes sign. 7. The first row of the matrix is multiplied by 5. If a row in a matrix is multiplied by a scalar, then the determinant of the matrix is multiplied by that scalar. 9. A 4 is factored out of the second column and a 3 is factored out of the third column. If a column of a matrix is multiplied by a scalar, then the determinant of the matrix is multiplied by that scalar. 11. The matrix is multiplied by 5. If an n n matrix is multiplied by a scalar c, then the determinant of the matrix is multiplied by c n. 13. 4 times the first row is added to the second row. If a scalar multiple of one row of a matrix is added to another row, then the determinant of the matrix is unchanged. 15. A multiple of the first row is added to the second row. If a scalar multiple of one row is added to another row, then the determinants are equal. 17. The second row of the matrix is multiplied by 1. If a row of a matrix is multiplied by a scalar, then the determinant is multiplied by that scalar. 19. The sixth column is 2 times the first column. If one column of a matrix is a multiple of another column, then the determinant of the matrix is zero. 21. 1 23. 19 25. 28 27. 17 29. 60 31. 223 33. 1344 35. 136 37. 1100 39. (a) True. See Theorem 3.3, part 1, page 134. (b) True. See Theorem 3.3, part 3, page 134. (c) True. See Theorem 3.4, part 2, page 136. 41. k 43. 1 45. 1 47. Proof 49. cos 2 sin2 1 51. sin2 1 cos2 53. Not possible. The determinant is equal to cos2 x sin2 x, which cannot equal zero because cos 2 x sin2 x 1. 55. Proof Section 3.3 1. (a) 0 (c) 2 4 (b) (page 149) 1 3 6 3. (a) 2 (c) (d) 2 0 3 4 2 1 8 5 1 1 0 12 (b) 4 0 2 6 3 3 0 (d) 0 5. (a) 3 (b) 6 6 3 2 1 (c) 9 4 8 5 (d) 18 7. 44 9. 54 11. (a) 2 (b) 2 (c) 0 13. (a) 0 (b) 1 (c) 15 15. (a) 14 17. (a) 29 19. (a) 22 (b) 196 (b) 841 (b) 22 (c) 196 (c) 841 (c) 484 (d) 56 (d) 232 (d) 88 1 1 1 (e) 14 (e) 29 (e) 22 21. (a) 115 23. (a) 15 25. (a) 8 (b) 115 (b) 125 (b) 4 (c) 13,225 (c) 243 (c) 64 (d) 1840 (d) 15 (d) 8 1 1 1 (e) 115 (e) 5 (e) 2 27. Singular 29. Nonsingular 31. Nonsingular 33. Singular 1 1 1 35. 5 37. 2 39. 24 41. The solution is not unique because the determinant of the coefficient matrix is zero. 43. The solution is unique because the determinant of the coefficient matrix is nonzero. 1, 4 45. k 47. k 24 49. Proof 0 1 1 0 51. and 53. 0 0 0 0 0 (The answer is not unique.) 55. Proof A24 Answer Key 57. No; in general, P 1AP A. For example, let 1 2 5 2 , P 1 , P 3 5 3 1 2 1 . and A 1 0 Then you have 27 49 P 1AP A. 16 29 A is true in general because The equation P 1AP P 1 AP P 1 A P 1 P 1 P A P A A. P 59. (a) False. See Theorem 3.6, page 144. (b) True. See Theorem 3.8, page 146. (c) True. See "Equivalent Conditions for a Nonsingular Matrix," parts 1 and 2, page 147. 61. Proof 63. Orthogonal 65. Not orthogonal 67. Orthogonal 69. Proof 2 3 2 3 1 3 2 3 1 3 2 3 1 3 2 3 2 3 2 3 2 3 1 3 2 3 1 3 2 3 1 3 2 3 2 3 5. (a) (c) 7. (a) (c) 9. (a) (c) 11. (a) (c) 2 2 2 3 0 (b) 2, 1 5 1 2: x ; 1: x 2 1 2 3 0 (b) 3, 1 1 1 3: x ; 1: x 1 3 3 18 0 (b) 6, 3 1 4 6: x ; 3: x 2 1 2 3 4 12 0 (b) 2, 3, 1 1 2: x 3: x 0 ; 1 ; 1 1 1 1 2: x 4 3 2 2 2 13. (a) (c) 3 3 0 1 ; 2 1 0 2 3, 1 3: x 1: x 0 (b) 1: x 1, 1, 3 1 0 2 71. (a) (b) (c) 1 1: x A is orthogonal. 73. Proof 3: x 15. Eigenvalues: Eigenvectors: 1 2 Section 3.4 1 0 1 0 1 3. 0 1 1 0 1 1 0 1 1. 2 3 2 3 1 1 1 1 1 1 1 1 1 (page 157) 1 0 1 1 2 1 1 0 0 1 1 1 1 0 0 1 1 1 1 0 1 1 1 1 ; 0 3 6 2 0 2 0 0 0 1 1 1 1 1 5 1 3 1 2 0 1 0 17. Eigenvalues: Eigenvectors: 1: x 1, 2, 3 0 1 ; 1 1 0 ; 1 2 0 1 1 0 1 1 1 1 1 2: x 3: x Answer Key 7 3 2 3 2 3 A25 13 3 5 3 2 3 19. Eigenvalues: Eigenvectors: 1: x 1 0 ; 0 1, 2 5. adj A 2: x 1 0 3 7. adj A 0 1 ; 0 0 7 2 2 7 7 4 2 7 9 12 3 3 1 1 2 1 1 9 1 9 2 9 1 9 13 5 ,A 2 9 0 9 9 1 0 1 1 0 1 4 1 1 1 21. Eigenvalues: Eigenvectors: 3: x 0 0 ; 1 1 1 0 ; 0 0 3, 1, 3, 5 13 4 , 10 5 13 9 4 9 10 9 5 9 1: x A 1 3: x 5: x 0 0 5 3 7 9 4 9 2 9 9. Proof 13. adj A 2 1 1 1 0 2 11. Proof 2, 1 23. Eigenvalues: Eigenvectors: 0 1 ; 0 0 1 0 ; 0 0 2, 1, 1, 3 A 1: x 1 0 2 6 1 0 2 2 2 1 2 2 2 2 21. x1 x2 3 4 1 2 2: x 15. Proof 17. x1 1 x2 2 19. x1 x2 1: x 3: x 25. (a) False. If x is an eigenvector corresponding to , then any nonzero multiple of x is also an eigenvector corresponding to . See page 153, first paragraph. a is an eigenvalue of the matrix A, then (b) False. If a is a solution of the characteristic equation I A 0. See page 153, second paragraph. 23. Cramer's Rule does not apply because the coefficient matrix has a determinant of zero. 25. Cramer's Rule does not apply because the coefficient matrix has a determinant of zero. 0 27. x1 1 29. x1 1 31. x1 1 1 x2 1 x2 2 x2 2 3 1 x3 x3 2 x3 2 2 33. Cramer's Rule does not apply because the coefficient matrix has a determinant of zero. 4 1 35. x1 37. x1 39. x1 7 41. x1 5 4k 3 4k 1 x , y 2k 1 2k 1 1 The system will be inconsistent if k 2. 3 47. 3 Collinear 51. Not collinear 3y 4x 0 55. x 2 Section 3.5 1. adj A 0 0 0 (page 168) 4 3 0 12 4 2 , A 1 1 2 3 2 1 1 2 43. 45. 49. 53. 3. adj A 0 6 ,A 2 1 does not exist. A26 1 Answer Key 57. 3 59. 2 61. Not coplanar 63. Coplanar 65. 4x 10y 3z 27 67. x y z 0 69. Incorrect. The numerator and denominator should be interchanged. 71. Correct 73. (a) 49a 7b c 4380 64a 8b c 4439 81a 9b c 4524 136, c 4695 (b) a 13, b (c) 4600 43. (a) False. See "Definitions of Minors and Cofactors of a Matrix," page 124. (b) False. See Theorem 3.3, part 1, page 134. (c) False. See Theorem 3.9, page 148. 45. 128 47. Proof 49. 0 51. 53. 7: x 1: x 1 and 1 1 3: x 1 ; 0 4: x 0 0 1 8: x 0 1 ; 0 2 1 and 0 4300 10 (d) The polynomial fits the data exactly. Review Exercises Chapter 3 (page 171) 1. 10 3. 0 5. 0 7. 6 9. 1620 11. 82 13. 64 15. 1 17. 1 19. Because the second row is a multiple of the first row, the determinant is zero. 21. A 4 has been factored out of the second column and a 3 has been factored out of the third column. If a column of a matrix is multiplied by a scalar, then the determinant of the matrix is also multiplied by that scalar. 1 2 23. (a) 1 (b) 5 (c) (d) 5 2 1 25. (a) 27. (a) 1 29. 6 33. x1 x2 x3 55. 1 57. uv 2 59. Row reduction is generally preferred for matrices with few zeros. For a matrix with many zeros, it is often easier to expand along a row or column having many zeros. 1 1 61. 63. Unique solution: x 0.6 2 0 y 0.5 65. Unique solution: x1 x2 x3 67. (a) 100a 400a 900a (c) 400 10b 20b 30b c c c 1 2 1 3 1 308.9 335.8 363.6 (b) a b c 0.0045 2.555 282.9 12 20 0 1 2 1 2 (b) (b) 1728 1 20 (c) 144 31. 10 35. x1 x2 x3 1 (d) 300 0 250 40 (d) The polynomial fits the data exactly. 3 1 2 69. 16 71. x 2y 4 73. 9x 4y 3z 0 75. (a) False. See Theorem 3.11, page 163. (b) False. See "Test for Collinear Points in the xy-Plane," page 165. 37. Unique solution 41. Not a unique solution 39. Unique solution Answer Key A27 Cumulative Test Chapters 13 1. 2. 3. 5. (page 177) 3, x3 2 x1 2, x2 x1 s 2t, x2 2 t, x3 t, x4 s x1 2s, x2 s, x3 2t, x4 t 4. k 12 BA 13,275.00 15,500.00 The entries represent the total values (in dollars) of the products sent to the two warehouses. 3, y 17 22 27 1 4 1 6 1 8 1 12 19. 1, 5 1 ; 1 2, 21. Proof 2 0 ; 1 2, 22. Proof 2 2 1 20. No; proof 23. (a) A is row-equivalent to B if there exist elementary matrices, E1, . . . , Ek , such that A Ek . . . E1B. (b) Proof 6. x 7. A A T 4 22 29 36 27 36 45 (b) 2 7 1 7 1 7 2 21 Chapter 4 Section 4.1 (page 188) 1. v 3. 1 x -1 -1 -2 -3 -4 -5 1 2 3 4 5 -5 -4 - 3 - 2 -1 -2 -3 1 4, 5 y 5. 1 y 8. (a) x 1 9. 0 3 5 0 0 1 5 1 1 9 5 (2, -4) (-3, -4) -5 -4 0 1 1 0 1 0 11. 34 1 0 2 1 0 4 (The answer is not unique.) 1 12. (a) 14 (b) 10 (c) 140 (d) 14 1 13. (a) 567 (b) 7 (c) 7 (d) 343 10. 4 11 1 11 2 11 7 2 6x 10 11 3 11 5 11 1 6x 7 11 1 11 2 11 7. u 5 4 3 2 1 -1 -1 v y 3, 1 9. u v 1, y 4 1 x - 5 - 4 -3 - 2 - 1 1 (3, 1) x 1 2 3 4 5 14. 15. a 1, b 0, c 2 (The answer is not unique.) 1 (2, 6) (-1, -4) 16. y 11. v 3, 9 2 y 13. v 5 4 3 2 8, 1 y y 6 5 4 3 2 (-3, ( 9 2 (-2, 3) 4 (-8, -1) v = u + 2w x -2 v= 3 u 2 u 2 x (-2, 3) (0, 1) (- 1, 2) -3 -2 -1 u x 2 3 2w -2 1 - 5 - 4 - 3 - 2 -1 (-6, -4) -4 17. 3x 2y 11 18. 16 Short Long Larson Texts, Inc. Final pages Elementary Linear Algebra 6e CYAN MAGENTA YELLOW BLACK A28 Answer Key 9 7 2, 2 15. v (c) y z 2 1 2v )- 9 , 7 ) 2 2 v= 1 2 (3u (-6, 9) 8 ( 1 , 1, 1( 2 (1, 2, 2) v 1 2 y 1 3u 4 2 x x 1 + w) w -8 (-3, - 2) -4 17. (a) 4 3 2 1 -1 -1 y (b) 4 y (4, 2) 2v v 1 2 2 v -4 (2, 1) x 2 (2, 1) x 3 4 -3v - 2 (-6, -3) -4 27. (a) and (b) 29. (a) 4, 2, 8, 1 (b) 8, 12, 24, 34 (c) 4, 4, 13, 3 31. (a) 9, 3, 2, 3, 6 (b) 2, 18, 12, 18, 36 (c) 11, 6, 4, 6, 3 33. (a) 1, 6, 5, 3 (b) 3 13 21 (c) (d) 2 , 11, 2, 2 35. 39. v 43. v 47. v 1 2, 7 2, 9 2, 1 4, 1, 3, 8, 10, 0 3, 1 2 2w 6, 3 2 2 37. 4, 8, 18, 41. v 45. 3u3 u 1, 5 3, (c) 2 y u u u1 w 2u2 1 1 v 2 (2, 1) v )1, 1 ) 2 1 2 x 19. u v v u 1, 0, 4 1, 0, 4 23. (b) 7 2, 21. 6, 12, 6 25. (a) 5 4 3 2 3, 5 2 z 2 z (1, 2, 2) 5 x 4 3 2 (2, 4, 4) v 2v 2 3 4 5 y 1 2 v (1, 2, 2) 1 2 -v (-1, -2, -2) x y 49. It is not possible to write v as a linear combination of u1, u2, and u3. 51. v 2u1 u2 2u3 u4 u5 53. v 5u1 u 2 u 3 2u4 5u5 3u6 55. (a) True. Two vectors in Rn are equal if and only if their corresponding components are equal, that is, u v if and only if u1 v1, u2 v2, . . . , u u vu . (b) False. The vector cv is c times as long as v and has the same direction as v if c is positive and the opposite direction if c is negative. 57. No 59. Answers will vary. 61. Proof 63. (a) Add v to both sides. (b) Associative property and additive identity (c) Additive inverse (d) Commutative property (e) Additive identity Answer Key A29 65. (a) Additive identity (b) Distributive property (c) Add c 0 to both sides. (d) Additive inverse and associative property (e) Additive inverse (f ) Additive identity 67. (a) Additive inverse (b) Transitive property (c) Add v to both sides. (d) Associative property (e) Additive inverse (f ) Additive identity 69. No 71. You could describe vector subtraction as follows: v u-v 23. The set is a vector space. 25. The set is not a vector space. Axiom 1 fails because 1 0 0 0 1 0 , 0 0 0 1 0 1 which is not singular. The set is a vector space. (a) The set is not a vector space. Axiom 8 fails because 1 2 1, 1 3 1, 1 3, 1 1 1, 1 2 1, 1 1, 1 2, 1 3, 2 . (b) The set is not a vector space. Axiom 2 fails because 1, 2 2, 1 1, 0 2, 1 1, 2 2, 0 . (Axioms 4, 5, and 8 also fail.) (c) The set is not a vector space. Axiom 6 fails because 1 1, 1 1, 1 , which is not in R 2. (Axioms 8 and 9 also fail.) Proof The set is not a vector space. Axiom 5 fails because 1, 1 is the additive identity so 0, 0 has no additive inverse. (Axioms 7 and 8 also fail.) (a) True. See page 191. (b) False. See Example 6, page 195. (c) False. With standard operations on R2, the additive inverse axiom is not satisfied. (a) Add w to both sides. (b) Associative property (c) Additive inverse (d) Additive identity Proof 41. Proof 27. 29. u 31. 33. or write subtraction in terms of addition, u v u 1 v. 35. Section 4.2 (page 197) 1. 0, 0, 0, 0 5. 0 7. 9. 0x a11 a21 0x 2 a12 a22 0x 3 v1, a13 a23 v2, a11 a21 v3, a12 a22 v4 a13 a23 39. 3. 0 0 0 0 0 0 37. v1, v2, v3, v4 a0 a1x a2x2 a3x3 a0 a1x a2x2 a3x3 11. 13. The set is a vector space. 15. The set is not a vector space. Axiom 1 fails because x3 1 1, which is not a third-degree polyx3 nomial. (Axioms 4, 5, and 6 also fail.) 17. The set is not a vector space. Axiom 4 fails. 19. The set is not a vector space. Axiom 6 fails because 1 x, y x, y , which is not in the set when x 0. 21. The set is a vector space. Section 4.3 (page 205) 1. Because W is nonempty and W R4, you need only check that W is closed under addition and scalar multiplication. Given and x1, x2, x3, 0 W y1, y2, y3, 0 W, it follows that x1, x2, x3, 0 y1, y2, y3, 0 x1 y1, x2 y2, x3 y3, 0 W. W, it So, for any real number c and x1, x2, x3, 0 follows that cx1, cx2, cx3, 0 W. c x1, x2, x3, 0 A30 Answer Key 3. Because W is nonempty and W M2,2, you need only check that W is closed under addition and scalar multiplication. Given 0 a1 0 a2 W W, and 0 0 b1 b2 it follows that 0 a1 0 a2 0 a1 a2 W. 0 0 0 b1 b2 b1 b2 So, for any real number c and 0 a W, it follows that b 0 0 a 0 ca W. b 0 cb 0 5. Recall from calculus that continuity implies integrability; W V. So, because W is nonempty, you need only check that W is closed under addition and scalar multiplication. Given continuous functions f, g W, it follows that f g is continuous and f g W. Also, for any real number c and for a continuous function f W, cf is continuous. So, cf W. c 7. Not closed under addition: 0, 0, 1 0, 0, 1 0, 0, 2 Not closed under scalar multiplication: 2 0, 0, 1 0, 0, 2 9. Not closed under scalar multiplication: 2 1, 1 2, 2 11. Not closed under scalar multiplication: 13. Not closed under scalar multiplication: 2 1, 1, 1 2, 2, 2 15. Not closed under addition: 1 0 0 0 1 0 0 0 1 0 0 1 1 ex ex 31. W is a subspace of R 3. (W is nonempty and closed under addition and scalar multiplication.) 33. W is a subspace of R 3. (W is nonempty and closed under addition and scalar multiplication.) 35. W is not a subspace of R 3. Not closed under addition: 1, 1, 1 1, 1, 1 2, 2, 2 Not closed under scalar multiplication: 2 1, 1, 1 2, 2, 2 37. (a) True. See "Remark," page 199. (b) True. See Theorem 4.6, page 202. (c) False. There may be elements of W which are not elements of U , or vice-versa. 3947. Proof Section 4.4 (page 219) 1. (a) u cannot be written as a linear combination of the given vectors. 3 (b) v 1 2, 1, 3 4 2 5, 0, 4 (c) w 8 2, 1, 3 3 5, 0, 4 (d) z 2 2, 1, 3 5, 0, 4 7 5 3. (a) u 0 2, 12, 13 4 2, 0, 7 4 2, 4, 5 (b) v cannot be written as a linear combination of the given vectors. 1 1 (c) w 0 2, 12, 13 6 2, 0, 7 3 2, 4, 5 (d) z 4 2, 0, 7) 5 2, 4, 5 0 2, 12, 13 5. S spans R 2. 7. S spans R 2. 9. S does not span R 2. It spans a line in R 2. 11. S does not span R 2. It spans a line in R 2. 13. S does not span R2. It spans a line in R 2. 15. S spans R2. 17. S spans R 3. 19. S does not span R 3. It spans a plane in R 3. 21. S does not span R 3. It spans a plane in R 3. 23. Linearly independent 25. Linearly dependent 27. Linearly independent 29. Linearly dependent 31. Linearly independent 33. Linearly independent 7 35. 3, 4 4 1, 1 0, 0 , 2 2, 0 3, 4 4 1, 1 2 2, 0 (The answer is not unique.) 7 17. Not closed under addition: 2, 8 3, 27 5, 35 Not closed under scalar multiplication: 2 3, 27 6, 54 19. Not a subspace 25. Subspace 21. Subspace 27. Subspace 23. Subspace 29. Not a subspace Answer Key A31 37. 1, 1, 1 1, 1, 0 0, 0, 1 1, 1, 1 1, 1, 0 0, 0, 1 (The answer is not unique.) 0 0, 1, 1 0 0, 1, 1 1 0, 0, 0 39. (a) All t 1, 2 (b) All t 2 6 19 3A 2B 41. (a) 10 7 (b) Not a linear combination of A and B 2 28 A 5B (c) 1 11 0 0 0A 0B (d) 0 0 43. Linearly dependent 45. Linearly independent 47. S does not span P2. 49. (a) Any set of three vectors in R 2 must be linearly dependent. (b) The second vector is a scalar multiple of the first vector. (c) The first vector is the zero vector. 51. S1 and S2 span the same subspace. 53. (a) False. See "Definition of Linear Dependence and Linear Independence," page 213. (b) True. See corollary to Theorem 4.8, page 218. 5561. Proof 63. The theorem requires that only one of the vectors be a 1, 0, 2 linear combination of the others. Because 0 1, 2, 3 1, 0, 2 , there is no contradiction. 65. Proof 67. On 0, 1 , f2 x x x 1 3 f1 x y 4 3 2 1, 1 , f1 and f2 are not multiples of each other. On f2 x , then c 3x x . But if For if they were, cf1 x 1 x 1, c 1, whereas if x 1, c . 3 3 69. Proof Section 4.5 (page 230) 1. R 6: 1, 0, 0, 0, 0, 0 , 0, 1, 0, 0, 0, 0 , 0, 0, 1, 0, 0, 0 , 0, 0, 0, 1, 0, 0 , 0, 0, 0, 0, 1, 0 , 0, 0, 0, 0, 0, 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 , 0 0 0 , 0 0 , 0 0 , 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 , 0 1 , 0 0 , 0 0 1 3. M2,4: 5. P4: 1, x, x2, x3, x4 9. 11. 13. 15. 17. 19. 21. 25. 27. 29. 31. 33. 35. 39. 41. 45. 49. 7. S is linearly dependent. 1 3 3x f1, f2 dependent f1 (x) = 3x f2(x) = |x| x 1 2 3 4 -4 -2 S is linearly dependent and does not span R 2. S is linearly dependent and does not span R 2. S does not span R2. S is linearly dependent and does not span R 3. S does not span R3. S is linearly dependent and does not span R 3. S is linearly dependent. 23. S is linearly dependent. S does not span M2,2. S is linearly dependent and does not span M2,2. The set is a basis for R 2. The set is not a basis for R 2. The set is not a basis for R 2. S is a basis for R 2. 37. S is a basis for R 3. 3. S is not a basis for R S is a basis for R 4. 43. S is a basis for M2,2. S is a basis for P3. 47. S is not a basis for P3. 3. S is a basis for R 8, 3, 8 2 4, 3, 2 0, 3, 2 3 0, 0, 2 51. S is not a basis for R3. 53. S is not a basis for R 3. 55. 6 57. 1 59. 8 61. 6 A32 Answer Key 1 0 0 0 0 0 0 0 0 0 0 , 0 1 0 , 0 0 0 , 63. 0 0 0 0 0 0 0 0 0 1 dim D3,3 3 65. 1, 0 , 0, 1 , 1, 0 , 1, 1 , 0, 1 , 1, 1 67. 1, 1 , 1, 0 (The answer is not unique.) 69. (a) Line through the origin (b) 2, 1 (c) 1 71. (a) Line through the origin (b) 2, 1, 1 (c) 1 73. (a) 2, 1, 0, 1 , 1, 0, 1, 0 (b) 2 75. (a) 0, 6, 1, 1 (b) 1 77. (a) False. If the dimension of V is n, then every spanning set of V must have at least n vectors. (b) True. Find a set of n basis vectors in V that will span V and add any other vector. 79. Proof 81. Proof 83. (a) Basis for S1: 1, 0, 0 , 1, 1, 0 , dimension 2 Basis for S2: 0, 0, 1 , 0, 1, 0 , dimension 2 Basis for S1 S2: 0, 1, 0 , dimension 1 Basis for S1 S2: 1, 0, 0 , 0, 1, 0 , 0, 0, 1 , dimension 3 (b) No, it is not possible. 85. Proof 9. (a) 2 (c) (b) 1 0 0 , 1 19 7 8 7 1, 2, 2, 0 , 0, 0, 0, 1 11. (a) 5 (b) 1, 0, 0, 0, 0 , 0, 1, 0, 0, 0 , 0, 0, 1, 0, 0 , 0, 0, 0, 1, 0 , 0, 0, 0, 0, 1 1 0 0 0 1 0 1 0 0 0 (c) 0 , 0 , 1 , 0 , 0 0 0 0 1 0 0 0 0 0 1 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 1, 0, 0 , 0, 1, 0 , 0, 0, 1 1, 1, 0 , 0, 0, 1 1, 0, 1, 0 , 0, 1, 0, 0 , 0, 0, 0, 1 1, 0, 0, 0 , 0, 1, 0, 0 , 0, 0, 1, 0 , 0, 0, 0, 1 0, 0 , dim 0 2, 1, 0 , 3, 0, 1 , dim 3, 0, 1 , dim 1 2 1, 2, 1 , dim 1 2, 2, 0, 1 , 1, 1, 1, 0 , dim 0, 0, 0, 0 , dim 0 (a) (b) 1 1, 3, 2 (a) (b) 2 3, 0, 1 , 2, 1, 0 2 4, 1, 1, 0 , 3, 3, 0, 1 (a) 2 Section 4.6 (page 246) 1. (a) 2 (c) 3. (a) 1 5. (a) 2 (c) 7. (a) 2 (c) (b) 1 0 , 0 1 (b) (b) 1 0 , 0 1 (b) 1 0 0 , 1 2 5 3 5 1, 0 , 0, 1 (b) 2 1, 2, 3 1 2 (c) 1 2 1 (a) 8, 9, 6, 6 (b) 1 (a) Consistent 3, 5, 0 (b) x t 2, 4, 1 43. (a) Inconsistent (b) Not applicable 45. (a) Consistent (b) x t 5, 0, 6, 4, 1 1, 0, 2, 3, 0 47. 1 4 5 4 1, 0, , 0, 1, s 2, 1, 0, 0, 0 1, 0, 1 , 0, 1, 3 4 2 2 1 1 2 2 0 3 4 3 4 3 1 0 1 2 49. 0 0 1 1 2 3 Answer Key A33 51. Four vectors in R 3 must be linearly dependent. 53. Proof 55. (a) 1 0 1 (b) 0 1 (c) 0 0 0 , 1 1 0 0 , 0 0 0 0 , 0 0 1 0 1 0 0 1 (e) R m Section 4.7 (page 260) 1. 8 3 1 1 2 5 3. 4 3 11. 0 1 2 1 15. 2 4 9 5 8 5 5. 1 2 0 1 13. 2 1 2 1 3 7. 3 2 3 2 1 2 9. 2 1 2 1 57. (a) m (b) r (c) r (d) R n 59. Answers will vary. 61. Proof 63. (a) True. See Theorem 4.13, page 233. (b) False. See Theorem 4.17, page 241. 65. (a) True. the columns of A become the rows of AT, so the columns of A span the same space as the rows of AT. (b) False. The elementary row operations on A do not change linear dependency relationships of the columns of A but may change the column space of A. 3 67. (a) rank A rank B n r 5 3 2 nullity A (b) Choosing x3 and x5 as the free variables, x1 s t x2 2s 3t x3 s x4 5t x5 t. A basis for the nullspace is 1 1 2 3 1 , 0 . 0 5 0 1 (c) 1, 0, 1, 0, 1 , 0, 1, 2, 0, 3 , 0, 0, 0, 1, 2, 1, 3, 1 , 5, 3, 11, 7 , 0, 1, 7, 5 (d) (e) Linearly dependent (f) (i) and (iii) 69. Proof 71. Proof 1 3 4 5 3 5 17. 0 0 0 1 12 19. 1 21. 3 3 24 10 29 12 1 0 27. 5 4 3 4 1 5 6 7 3 7 3 3 11 2 11 9 22 1 2 1 11 1 3 3 4 1 4 5 1 0 3 1 5 11 3 22 19 44 1 4 2 11 1 3 1 2 23. 2 1 2 1 0 0 1 4 1 4 7 11 1 11 21 22 1 2 5 11 7 5 11 3 1 3 10 6 10 25. 0 29. (a) 0 (b) (d) 6 9 6 3 1 2 4 4 (c) Verify. 5 31. (a) 4 7 2 5 10 2 1 1 0 1 2 1 2 1 2 (b) 1 2 3 2 5 4 3 4 5 4 A34 Answer Key 11 4 35. Parabola y 2 1 17 20 3 10 6 5 5 4 1 2 37. Ellipse y 5 4 3 x -1 -2 -4 -2 -3 -4 -5 1 x 1 2 4 5 (c) Verify. (d) 9 4 5 4 3 32 1 16 1 2 279 160 61 80 7 10 48 5 24 10 5 33. (a) 4 6 5 4 5 1 2 2 5 -4 -3 -2 -1 (b) 0 y2 + x = 0 x 2 + 4y 2 - 16 = 0 (c) Verify. (d) 39. Hyperbola y 6 41. Parabola y 1 -2 -1 x 4 35. 11 1 37. 1 5 2 0 39. 3 2 41. 1 2 1 4 x 2 -4 -6 x2 y2 - - 9 16 (1, - 2) 3 4 - 6 -4 4 6 -4 -5 43. (a) False. See Theorem 4.20, page 253. (b) True. See the discussion following Example 4, page 257. 45. QP 47. If B is the standard basis, then B B B I I B 1 shows that P 1, the transition matrix from B to B , is B 1. I B If B is the standard basis, then B B 1, shows that P the transition matrix from B to B , is B. 1=0 x 2 - 2x + 8y + 17 = 0 43. Point y 2 1 (2, 1) x 1 2 Section 4.8 (page 270) 1. 5. 11. 17. 21. 25. 27. 29. 33. (b), (c), and (d) 3. (a), (b), and (d) (b) 7. (b) 9. 2 x 13. 0 15. 2e3x Linearly independent 19. Linearly dependent Linearly dependent 23. Linearly dependent y C1 sin x C2 cos x y C1 C2 sin x C3 cos x Proof 31. Proof 1. Two solutions are No. For instance, consider y x2 x2 y 1. Their sum is not a solution. and y 2 2 9x 2 + 25y 2 - 36x - 50y + 61 = 0 45. Hyperbola (- 3, 5) 8 6 4 2 -6 -4 -2 x y 1 9x 2 - y 2 + 54x + 10y + 55 = 0 Answer Key A35 47. Ellipse y -5 -4 - 3 -2 - 1 -2 -4 x 49. Hyperbola y 61. y 4 x 2 63. y x y' 0 y 2 1 x' 10 8 y x' y' 60 2 -2 -4 x 4 6 (-1, 5) 2 -4 -2 2 4 45 -2 -1 -1 -2 1 2 x -6 x 2 + 4y 2 + 4x + 32y + 64 = 0 2x 2 - y 2 + 4x + 10y - 22 = 0 65. x 51. Parabola y 1 -4 -3 -2 -1 x -1 -2 -3 2 2 y 3 2 1 y' x' 45 1 2 3 - 3 - 2 -1 -1 -2 -3 x x 2 + 4x + 6y - 2 = 0 y 53. 2 y' 2 x 2 y 2 1 x' x 55. 3 y' 2 y 5 y 3 1 2 67. Proof 1 x' 69. Proof Review Exercises Chapter 4 (page 272) 1. (a) 0, 2, 5 (b) 2, 0, 4 2, 2, 1 (c) 5, 6, 5 (d) 5. 9. v 1 2, 1 45 -2 -1 x 1 -1 -2 -3 -1 45 x 1 3 3. (a) (b) (c) (d) 7. 5 2, 3, 1, 4, 4 0, 4, 4, 2 3, 3, 0, 2 9, 7, 2, 7 6, 0 9 8 u1 1 8 u2 4, 2u1 0 0 0 4 u2 0 0 0 a11 a21 a31 3u3 0 0 0 a12 a22 a32 11. v 0 0 , 0 a13 a23 a33 -3 0u3 57. x 6 y' 2 y 4 y 3 1 2 1 x' 59. x y' 2 y 4 y 2 2 13. O3,4 1 A x' a14 a24 a34 45 1 3 30 x -2 -2 x 2 15. O -3 -1 -2 -3 0, 0, 0 a1, a2, a3 A 17. W is a subspace of R 2. 19. W is not a subspace of R 2. 21. W is a subspace of R 3. A36 Answer Key 23. W is not a subspace of C 1, 1 . 25. (a) W is a subspace of R 3. (b) W is not a subspace of R 3. 27. (a) Yes (b) Yes (c) Yes 29. (a) No (b) No (c) No 31. 33. 35. 37. 39. 41. 43. (a) Yes (b) No (c) No S is a basis for P3. The set is not a basis for M2,2. 3, 0, 4, 1 , 2, 1, 0, 0 (a) (a) 2, 3, 7, 0 , 1, 0, 0, 1 8, 5 Rank A nullity A 1 1 2 1 (b) 2 (b) 2 2 4 3 3, 0, 1, 0 , 1, 2, 0, 1 Rank A nullity A 2 45. 4, 2, 1 Rank A nullity A 2 47. (a) 2 49. (a) 1 51. (a) 3 53. 2 8 (b) 1, 0 , 0, 1 (b) 1, 4, 0, 4 (b) 1, 0, 0 , 0, 1, 0 , 0, 0, 1 55. 3 4 1 4 Yes, W is a subspace of V. Proof Answers will vary. (a) True. See discussion above "Definitions of Vector Addition and Scalar Multiplication in R n," page 183. (b) False. See Theorem 4.3, part 2, page 186. (c) True. See "Definition of Vector Space" and the discussion following, page 191. 85. (a) True. See discussion under "Vectors in Rn," page 183. (b) False. See "Definition of Vector Space," part 4, page 191. (c) True. See discussion following "Summary of Important Vector Spaces," page 194. 87. (a) and (d) are solutions. 89. (a) is a solution. 91. e x 93. 8 95. Linearly independent 97. Linearly dependent 99. Circle 101. Hyperbola y 2 x 1 2 3 4 y 77. 79. 81. 83. 57. 3 1 63. 0 1 69. 2 1 1 3 2 (- 1, 0) -4 - 2 -1 -2 -3 x 59. 2 5 1 4 61. 1 4 3 2 2 -4 65. 0 71. 0 1 12 6 0 1 0 1 0 0 67. 2 1 2 x 2 + y 2 - 4x + 2y - 4 = 0 1 1 3 1 x 2 - y 2 + 2x - 3 = 0 103. Parabola y 50 40 30 20 10 x -10 1 2 3 4 6 7 8 105. Ellipse y 1 -6 -4 - 2 -1 -2 -3 -4 -5 x 73. Basis for W: x, x2, x3 Basis for U: x 1 , x x 1 , x2 x 1 Basis for W U: x x 1 , x2 x 1 75. No. For example, the set x 2 x, x 2 x, 1 is a basis for P2. 2x 2 (5, - 4) - 20x - y + 46 = 0 (- 4, -2) 4x 2 + y 2 + 32x + 4y + 63 = 0 Answer Key A37 107. x 6 y' 2 y 6 y 6 2 1 109. x 2 4y y 12 10 8 6 4 2 1 y' 43. (a) u 13, v 17 8 15 (b) 17 , 17 (c) (d) (e) (f) 47. (a) 5 12 13 , 13 45. (a) u 13, v 13 12 5 (b) 13 , 13 5 (c) 13 , (d) 0 (e) 169 (f) 169 5 (b) 13 , 12 13 x' 45 -6 -6 x 3 6 140 169 289 u 26, v 13 5 12 13 , 13 -6-4-2 -4 -36.87 x' x 12 13 Chapter 5 Section 5.1 1. 5 7. (a) 17 4 (page 290) 10 24 (c) 26 , 26 (e) 676 (d) 238 (f) 169 (b) 0, 5 13 , 12 13 49. (a) u 3. 3 (b) 5 41 8 6 11 13 5 38 5 38 1 3, 13, v 5 13 , 12 13 13 5. 3 6 (c) 577 8 13. (a) (b) 5 13 , 5 13 , 12 13 12 13 (c) 0, (e) 169 (d) 169 (f) 169 9. (a) 5 (b) 6 (c) 21 15. (a) (b) 17. (a) 19. 23. 1, 1 3, 11. (a) (b) (c) 2 , 38 , 2 38 , 1.0843, v 0.3202 51. (a) u (b) 0, 0.7809, 0.6247 0.9223, 0.1153, 0.3689 (c) (d) 0.1113 (e) 1.1756 (f) 0.1025 1.7321, v 2 53. (a) u 0.5, 0.7071, 0.5 (b) (c) 0, 0.5774, 0.8165 (d) 0 (e) 3 (f) 4 55. (a) u 3.7417, v 3.7417 3 , 38 3 38 0, 2 2 3, 3 (b) 0, 2 3, 2 3 1 14 3, 0 21. 2 2, 2 2 25. 0, 2, 6 3 , , 6 6 3 2 (b) 0.5345, 0, 0.2673, 0.2673, 0.5345, 0.5345 (c) 0, 0.5345, 0.5345, 0.2673, 0.2673, 0.5345 (d) 7 (e) 14 (f) 14 3 6 57. (a) u 3.7417, v 4 0.25, 0, 0.25, 0.5, 0.5, 0.25, 0.25, 0.5 (b) (c) 0.2673, 0.2673, 0.5345, 0.2673, 0.2673, 0.2673, 0.5345, 0.2673 (d) 4 (e) 14 (f) 16 2, 3 3, 4 2, 3 59. 3, 4 6 5 13 1, 3, 2 61. 1, 1, 2 1, 1, 2 1, 3, 2 2 2 21 7 63. 1.713 radians; 98.13 65. ; 105 12 2, 27. (a) 4, 4, 3 (b) 16, 16, 12 (c) 29. 2 2 35. (a) 6 (b) 25 (c) 25 12, 18 (d) (e) 30 41. 7 31. 2 37. (a) (b) (c) (d) (e) 3 0 6 6 0 0 33. 22 39. (a) 5 (b) 50 (c) 50 (d) 0, 10, 25, 20 (e) 25 A38 Answer Key 67. 1.080 radians; 61.87 69. 75. 79. 81. 87. 93. 95. 71. 73. v t, 0 4 2 77. v v 2t, 3t t, 4t, s r, 0, s, t, w v Orthogonal 83. Parallel 85. Neither Neither 89. Orthogonal 91. Neither Orthogonal; u v 0 (a) False. See "Definition of Length of a Vector in R n," page 278. (b) False. See "Definition of Dot Product in Rn," page 282. (a) u v is meaningless because u v is a scalar. u v is meaningless because u is a vector (b) u and u v is a scalar. 5, 1 4, 0 1, 1 26 4 2 1, 1 1, 1 2, 0 2 2 2 2, 0 7, 1, 2 Section 5.2 1. (a) 33 (b) 5 (c) 13 (d) 2 65 7. (a) 0 9. (a) 3 11. (a) (page 303) 3. (a) 15 (b) 57 (c) 5 (d) 2 13 (c) (c) 3 411 (d) 3 5. (a) (b) (c) (d) 34 97 101 266 (b) 8 3 (b) 6 (d) 3 67 97. 16 10 2 210 (b) (c) 15 5 15 6 2 13. (a) (b) 0.736 0.816 e 3 1 e2 (c) 1.904 2 2e2 (d) 15. (a) 0 17. (a) 19. (a) 21. (a) 6 5 4 e2 2 2 3 (b) (b) (b) (b) 1 2e2 2 35 39 11 2 4 e (c) 1.680 (d) (d) 2 99. 101. 103. 105. 4 2 2 1, 1 2 2 54 2 1, 1 2 2 2 2 2 2 3, 4, 29 4, 3, 0 2 52 0 and v 2 < 15 8 15 8 107. (a) (b) 0 2 provided u < 0 2 (c) 15, 8 , and 17, 17 and 109. 15, 8 and 17 , 17 are orthogonal to 8, 15 . The answer is not unique. 111. cos 1 6 3 35.26 113117. Proof 119. Ax 0 means that the dot product of each row of A with the column vector x is zero. So, x is orthogonal to the row vectors of A. 121. Proof 23. (a) 0 (b) 25. Proof 27. Proof 29. Axiom 4 fails, page 293. 0, 1 , 0, 1 0, but 0, 1 0. 31. Axiom 4 fails, page 293. 1, 1 , 1, 1 0, but 1, 1 0. 33. Axiom 2 fails, page 293. 1, 0 , 1, 0 1, 0 1, 0 , 2, 0) 4 1, 0 , 1, 0 1, 0 , 1, 0 1 1 2 1, 2 and v 2, 3 , 35. Axiom 1 fails, page 293. If u 22 5 and u, v 31 3 v, u 32 2 31 9. 37. 2.103 radians 120.5 39. 1.16 radians 66.59 41. 45. 2 2 43. 1.23 radians 70.53 2 2 5 (c) 7 (c) 5 (c) 2 (c) 2 3 2 5 (d) 3 6 (d) 3 6 (d) 21 (d) 2 Answer Key A39 47. (a) (b) 49. (a) (b) 51. (a) (b) 53. (a) (b) 55. (a) (b) 5, 12 (3, 4 13 5 63 5, 12 3, 4 5, 12 3, 4 8 5 13 5 1, 0, 4 5, 4, 1 17 42 1 714 4, 4, 5 17 42 57 17 42 2 2x, 3x 1 2x 3x2 1 0 2 10 2 2x 3x 1 2x 3x2 1 14 2 10 0 3 31 24 13 14 35 14 14 35 3 4 14 35 6 4 77 14 35 sin x, cos x sin x cos x sin x 0 cos x 2 x ex 1 3 5, 12 , 3, 4 63. (a) (c) 8 4 5, 5 (b) 4 8 5, 5 y 2 u = (1, 2) v = (2, 1) 1 projuv projvu 1 2 4 12 5, 5 x 65. (a) 1, 1 (c) 5 (b) y (4, 4) v (- 1, 3) u 3 projuv -1 5 67. (a) 0, 2, 5 2 1 2, projvu 1 2 3 4 5 14 , x 15 5 14 , 7 5 46 , 15 15 46 , 23 (b) 1, 1 69. (a) sin x cos x 1 2, (b) 0, 71. projg f 75. projg f 1 2 2e 1 2 0 0 73. projg f 77. projg f 2ex e 2 57. (a) x, ex (b) 1 x 11 6 1 sin 2x ex 1 2 2e x 1 3 ex 1 2 2e 1 2 59. Because f, g cos x sin x dx 79. (a) False. See the introduction to this section, page 292. (b) True. See paragraph after "Remark," page 301. 42 22 2 0 u and v are 81. (a) u, v orthogonal. y (b) u = (4, 2) 2 1 x -1 -2 1 2 3 4 1 2 0, sin x 2 f and g are orthogonal. 1 3 x and g x 3x are 61. The functions f x 2 5x orthogonal because 1 1 f, g x 5x3 3x dx 1 2 1 1 1 1 5 5x 4 3x 2 dx x x3 0. 2 1 2 1 v = (2, - 2) Not orthogonal in the Euclidean sense 8389. Proof A40 Answer Key Section 5.3 1. Orthogonal 7. Orthogonal 13. Orthonormal 15. Orthogonal; (page 318) 3. Neither 9. Orthonormal 5. Orthonormal 11. Orthogonal 47. (a) True. See "Definitions of Orthogonal and Orthonormal Sets," page 306. (b) True. See corollary to Theorem 5.10, page 310. (c) False. See "Remark," page 314. 49. 51. 53. 2 5 3 10 10 5 , 0, , 0 , 0, , 0, 10 10 5 5 2 2 6 6 6 , 0, ,0 , , 0, , 2 2 6 6 3 10 3 190 9 190 3 10 , ,0 , , , 10 10 190 190 57. x2, x, 1 1 6 x2 2x 1 1 190 19 1 4 1 4 , , , 17 17 17 17 3 3 3 2 2 , 17. Orthogonal; , , , 0, 3 3 3 2 2 19. The set 1, x, x 2, x3 is orthogonal because 1, x 0, 1, x 2 0, 1, x3 0, 2 3 x, x 0, x, x 0, x 2, x3 0. Furthermore, the set is orthonormal because 1 1, x 1, x 2 1, and x3 1. 2, x3 is an orthonormal basis for P . So, 1, x, x 3 4 13 13 7 13 13 27. 31. 33. 35. 37. 39. 41. 3 4 4 5, 5 , 5, 1 2 2 3, 3, 3 , 4 3 5, 5, 0 , 3 5 2 2 1 2 3, 3, 3 , 3, 3 4 5 , 5 , 0 , 0, 55. Orthonormal 59. 2 2 1 61. , , 3 3 63. Proof 67. x2 1, 21. 23. 10 2 2 10 2 29. 1 3, 2 3 2 2 2 , 6 3 65. Proof 1 1 , 0, , 2 2 11 25. 2 15 0, 1 , 1, 0 1 1 , 0, , 0 , 0, 2 2 1 , 0, 2 1 1 1 , 0 , 0, , 0, 2 2 2 0, 1 6 , 6 3 , 3 3 3 , 3 3 0, 2 2 , , 2 2 6 6 , , 3 6 4 2 3 2 5 2 , , 7 14 14 3 4 5, 5, 0 , 4, 5 3 5, 0 3 3 3 , , 0, , 3 3 3 6 6 , , 6 3 3 , 3 3 , 3 1 6 ,0 , 6 3 ,0 3 x2 2 1 69. N A basis: 3, 1, 2 N AT basis: 1, 1, 1 R A basis: 1, 0, 1 , 1, 2, 3 R AT basis: 1, 1, 1 , 0, 2, 1 71. N A basis: 1, 0, 1 N AT 0, 0 R A basis: 1, 1 , (0, 1 } R2 R AT basis: 1, 0, 1 , 1, 1, 1 73. Proof Section 5.4 0 1 0 (page 333) 1. Not orthogonal 3. Orthogonal 0 0 , 1 0 2 1 0 1 43. x, 1 1 x dx 1 0 1 1 1 5. span 2 3 7. span 45. x2, 1 1 x2 dx x3 3 Answer Key A41 9. span 1 0 2 1 , 0 0 0 1 0 11. 2 3 2 3 2 3 5 3 13. 8 3 13 3 15. 1 1 2 37. Advanced Auto Parts: S 2.859t 3 32.81t 2 492.0t 2234 2010: S $6732 million Auto Zone: S 8.444t 3 105.48t 2 578.6t 4444 2010: S $8126 million 39. y 36.02t2 87.0t 6425 41. y 2.416t 3 36.74t 2 4989.3t 28,549 Explanations will vary. 43. (a) False. See discussion after Example 2, page 322. (b) True. See "Definition of Direct Sum," page 323. (c) True. See discussion preceding Example 7, page 328. 45. Proof 47. Proof 49. If A has orthonormal columns, then ATA I and the normal equations become ATAx AT b x AT b. 17. N A basis: 3, 0, 1 T NA 0, 0 R A basis: 1, 0 , 2, 1 R2 T basis: RA 1, 2, 3 , 0, 1, 0 1, 1, 0, 1 , 0, 1, 1, 0 19. N A basis: N AT basis: 1, 1, 1, 0 , 1, 2, 0, 1 R A basis: 1, 0, 1, 1 , 0, 1, 1, 2 R AT basis: 1, 0, 0, 1 , 0, 1, 1, 1 21. x 1 1 y Section 5.5 1 3 (page 350) 23. x 29. 2 2 1 1. j i z k 3. j k i z 25. x y 2 0 1 3 3 x 5 27. y = - x + 1 3 (- 1, 1) -3 -2 -1 3 y=1 + 4 (2, 2) 2 y (1, 0) 2 3 -2 -3 1 i x y x (- 1, 0) -2 (1, 0) 1 2 x x -k (3, - 3) y (-2, - 1) -1 -2 5. i k z j 31. 3 (-1, 2) 2 1 (1, 2) y= 7 5 -j y (- 2, 1) -2 -1 -1 (0, 1) (2, 1) x 1 2 x 33. y x2 x 35. y 3 2 7x 6 5x 26 35 7. 11. 15. 2, 1, 2, 3, 2, 1, 1 1 1 8, 14, 54 9. 1, 12, 2 13. 17. 5, 4, 3 A42 Answer Key 1, 5, 1 1 9 6 37. 2 51. (a) g x (b) 3 19. 23. 25. 31. 2, 1, 1 3 27. 6 5 33. 1 39. 2 4 135 21. 1, 1, 3 59. (a) g x (b) 1 0.0505 y 1.3122x 0.4177x2 f g 29. 2 83 35. Proof 4149. Proof 25 11x 0.4177x2 y x g f 0 0 4.5 61. (a) g x (b) 1 0.9802 53. (a) g x (b) 1 g 0.812x 0.8383 f - 2 g 2 x f 0 0 1 63. g x 65. g x 67. g x 69. g x 2 sin x 2 sin 2x 2 3 sin 3x 4 cos 3x 9 sin x 4 sin x 2 3 55. (a) g x (b) 1.5 0.1148 0.6644x g f 3 1 1 2 1 1 sin 2x e 20 4 cos x e 4 2 cos 2x 1 cos x 5 8 cos x 0 0 2 57. (a) g x (b) 1 1.5x2 y 0.6x f g 0.05 71. g x 73. g x 75. g x 2 sin x sin 2x 2 sin 2x sin 3x 3 sin 3x sin nx n 2 sin x . . . (page 352) Review Exercises Chapter 5 x 1 1. (a) 5 (b) 17 (c) 6 (d) 10 5. (a) 6 (b) 3 (c) 1 (d) 11 3. (a) 6 (b) 14 (c) 7 (d) 6 7. (a) 7 (b) 7 (c) 6 (d) 2 Answer Key A43 9. v 11. v 13. 19. 25. (a) 2 9 45 13 , 13 38; u 6; u 15. 21. 5 38 1 , 6 12 , 3 , 38 1 2 , 6 6 17. 23. 2 38 1 41. f, g 1 1 1 2x 1 1 x2 2x 1 x2 dx x4 4 1 x2 dx 1 2 1 x x3 dx 2 18 12 24 29 , 29 , 29 24 60 29 , 29 x2 2 1 0 1 2 3 11 (b) 2 1 3 2, 43. 1 27. Triangle Inequality: 2, 1 2, 2 2 6 (The answer is not unique.) , 0, , 1 , 2 6 , 1 6 2, 1 1 2, 2, 1 3 2, 2, 1 45. (a) 0 (b) Orthogonal 0, it follows that (c) Because f, g 4753. Proof 55. span 2 1 3 f, g f g . 67 15 53 2 2 4 Cauchy-Schwarz Inequality: 2, 1 2, 1 , 2, 2, 3 1 2, 1 2, 1 3 2, 2, 1 2 29. s, 3t, 4t 33. 35. 1 2 , 1 2 , 1 2 , 15 2 53 4 9.969 2s t, r, s, t 57. N A basis: 1, 0, 1 N AT 3, 1, 0 R A basis: 0, 0, 1 , 1, 3, 0 R AT basis: 0, 1, 0 , 1, 0, 1 59. y 1.778t 3 5.82t 2 77.9t 1603 2010: y $3578 million 61. y 10.61x 396.4 63. y 91.112x2 365.26x 315.0 65. y 151.692x2 542.62x 10,265.4 2, 1, 1 67. 69. i j 2k 71. Proof 75. g x y 31. 2r 1 2 3 5 0, 3, 4 , 1, 0, 0 , 0, 4, 5 5 5 1, 4, 2 1 0, , 2 1, 4, 2 2 0, 2, 1 2 , 37. (a) (b) (c) 2 1 3 , 1 2 , 3 2 0, 3 1, 0, 2 1 1 , 3 3 1 2 1 , 3 1 3 73. 2 18 x 5 8 5 f 77. g x y 3x 2 3 2 1 , 3 1 30 6 4 g 1 2 1 39. (a) 4 (d) 1 (b) 5 3 x, 5 4x2 (c) 3x 4 f g x 2 x -1 2 1 2 A44 Answer Key 79. y 4 0.3274x 2 1.459x 2.12 y f(x) 0 0 6 7. (a) A set of vectors v1, . . . , vn is linearly independent if the vector equation c1v1 . . . cn vn 0 has only the trivial solution. (b) Linearly dependent 8. The dimension is 6. One basis is 1 0 0 0 1 0 0 0 1 0 0 0 , 1 0 0 , 0 0 0 , 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 , 0 0 1 , 0 0 0 . 0 0 0 0 1 0 0 0 1 9. (a) A set of vectors v1, . . . , vn in a vector space V is a basis for V if the set is linearly independent and spans V. (b) Yes 1 0 4 1 0 6 10. 11. , 0 1 5 0 1 12. 0 2 1 1 0 1 1 1 1 81. g x 2 sin x sin 2x 83. (a) True. See note following Theorem 5.17, page 338. (b) True. See Theorem 3.18, page 339. (c) True. See discussion on pages 346 and 347. Cumulative Test--Chapters 4 and 5 1. (a) 3, 7 1 -1 -2 -3 -4 -5 -6 -7 -8 x y (page 359) v = (1, - 2) 7 8 w = (2, - 5) v + w = (3, - 7) y (b) 3, 6 1 -1 -2 -3 -4 -5 -6 -7 x v = (1, - 2) 6 7 13. (a) 5 (b) 11 (c) 4 (d) 1.0723 radians; 61.45 3v = (3, - 6) 14. 15. 11 12 (c) 6, 16 20 16 y 1, 0, 0 , 0, 1 13 2 2 , , 0, 2 2 2 2 , 2 2 2v - 4w = (- 6, 16) 16. 3, 2 y v = (- 3, 2) - 8 -4 -8 -12 -16 -20 2v = (2, - 4) x 2 1 u = (1, 2) 4w = (8, - 20) -3 -2 -1 x 1 2. w 3v1 4. Yes v2 2 3 v3 3. Proof 6. Yes projv u 1 = 13 (- 3, 2) -2 5. No Answer Key A45 17. N A basis: 0, 1, 1, 0 N AT 0, 0, 0 RA R3 R AT basis: 0, 1, 1, 0 , 18. span 1921. Proof 22. y y 3 2 1 -1 -2 -3 -4 -5 -6 -7 36 13 20 13 x 21. Linear 23. 3, 11, 5 2, 8 25. 0, 6, 8 1, 0, 0, 1 , 1, 1, 1, 1 1 1 1 27. 5, 0, 1 29. 2, 2 33. T: R 5 R 2 35. T: R 2 R 2 37. (a) 10, 12, 4 (b) 1, 0 (c) The system represented by 1 2 1 v1 2 4 1 v2 2 2 1 is inconsistent. 39. (a) 1, 1, 2, 1 (b) 1, 1, 1, 1 2 31. T: R4 R3 (1, 1) (2, 0) 1 2 3 4 5 6 7 8 9 x y = 2.7 - 1.5x (5, -5) 23. (a) 3 (b) One basis consists of the first three rows of A. (c) One basis consists of columns 1, 3, and 4 of A. 2 3 1 0 0 , 5 , (d) 0 1 0 1 0 0 (f) No (g) Yes 24. Proof 2 0 3 7 0 1 (h) No 41. (a) 0, 4 2 (b) 2 3 2, 2 3 2 5 5 3 (c) , 2 2 43. True. Dx is a linear transformation and preserves addition and scalar multiplication. 45. False, because sin 2x 2 sin x for all x. 47. g x 49. g x x2 x C cos x C 1 51. (a) 1 (b) 12 (c) 4 1 1 53. T 1, 0 55. x2 3x 5 2, 2 T 0, 2 1, 1 57. (a) True. See discussion before "Definition of a Linear Transformation," page 362. (b) False, because cos x1 x2 cos x1 cos x2. (c) True. See discussion following Example 10, page 370. 59. (a) x, 0 (b) Projection onto the x-axis (e) No 61. (a) 1 x y , 1 x 2 2 (c) Proof 63. Au 1 2 1 2 1 2 1 2 y (b) 5, 5 2 2 (d) Proof 1 2x 1 2x 1 2y 1 2y Chapter 6 Section 6.1 (page 371) 1. 3. 5. 7. 9. 15. (a) 1, 7 (a) 1, 5, 4 (a) 14, 7 (a) 0, 2, 1 Not linear Not linear (b) 11, 8 (b) 5, 6, t (b) 1, 1, t 6, 4 (b) 11. Linear 17. Linear x y Tu 13. Not linear 19. Linear 65. (a) Assume that h and k are not both zero. Because T 0, 0 0 h, 0 k h, k 0, 0 , it follows that T is not a linear transformation. 2, 1 (b) T 0, 0 T 2, 1 0, 0 3, 5 T 5, 4 A46 Answer Key (c) If T x, y x h, y k x, y , then h 0 and k 0, which contradicts the assumption that h and k are not both zero. Therefore, T has no fixed points. 67. Let T: R3 R3 be given by T x, y, z 0, 0, 0 . Then, if v1, v2, v3 is any set of vectors in R3, the set T v1 , T v2 , T v3 0, 0, 0 is linearly dependent. 6973. Proof 37. Nullity 2 Kernel: x, y, z : x 2y 2z 0 (plane) Range: t, 2t, 2t , t is real line) 39. 2 Zero 41. 4 Standard Basis 43. (a) 0, 0, 0, 0 0 0 (b) 0 0 0 0 0 0 0 (d) p x (e) 0, 0, 0, 0, 0 (c) Section 6.2 (page 385) 0, 0, 0, 0 a1x a2x a3x : a1, a2, a3 are real 9. 0, 0 a0: a0 is real (a) 0, 0 13. (a) 4, 2, 1 (b) 1, 0 , 0, 1 (b) 1, 0 , 0, 1 15. (a) 0, 0 (b) 1, 1, 0 , 0, 0, 1 2 3 1. 5. 7. 11. R3 3. 1, 0, 0, 0 , 0, 0, 1, 0 , 1 0 0 1 , , 0 0 0 0 0, 1, 0, 0 , 0, 0, 0, 1 0 0 0 0 , 1 0 0 1 17. (a) (b) 19. (a) 21. (a) (c) (d) 23. (a) (b) 25. (a) (c) 27. (a) (b) (c) (d) 1, 1, 1, 0 1, 0, 1, 0 , 0, 1, 0, 0 0, 0 4s, 4t, s 2 1 1, 0 , 0, 0, 0, 1 (d) 2 (b) 0 (c) R 2 (b) 0 t : s and t are real 11t, 6t, 4t : t is real (c) R2 (d) 2 t, 3t : t is real (b) 1 3t, t : t is real (d) 1 s t, s, 2t : s and t are real 2t, 2t, t : t is real 5s, s : s and t are real 20s 2t : r, s, and t are real 33. Nullity 3 Kernel: R 3 Range: 0, 0, 0 2 1 29. (a) 2s t, t, 4s, (b) 2 (c) 7r, 7s, 7t, 8r (d) 3 31. Nullity 1 Kernel: a line Range: a plane 35. Nullity 0 Kernel: 0, 0, 0 Range: R3 1 0 0 1 0 0 0 0 , , , 0 0 0 0 1 0 0 1 2, x3 1, x, x 1, 0, 0, 0, 0 , 0, 1, 0, 0, 0 , 0, 0, 1, 0, 0 , 0, 0, 0, 1, 0 a0 45. The set of constant functions: p x 47. (a) Rank 1, nullity 2 (b) 1, 0, 2 , 1, 2, 0 1 0, the homogeneous equation 49. Because A Ax 0 has only the trivial solution. So, ker T 0, 0 and T is one-to-one (by Theorem 6.6). Furthermore, because rank T dim R2 nullity T 2 0 2 dim R 2 T is onto (by Theorem 6.7). 1 0, the homogeneous equation 51. Because A Ax 0 has only the trivial solution. So, ker T 0, 0, 0 and T is one-to-one (by Theorem 6.6). Furthermore, because rank T dim R3 nullity T 3 0 3 dim R3 T is onto (by Theorem 6.7). 53. (a) False. See "Definition of Kernel of a Linear Transformation," page 374. (b) False. See Theorem 6.4, page 378. (c) True. See discussion before Theorem 6.6, page 382. (d) True. See discussion before "Definition of Isomor-phism," page 384. 55. (a) Rank n (b) Rank < n Answer Key A47 57. mn jk 59. Proof 61. Although they are not the same, they have the same dimension (4) and are isomorphic. 2 21. (a) 2 2 2 2 2 2 2 y 6 4 (b) 4 2,0 Section 6.3 (page 397) 1. 1 1 1 1 1 0 0 0 0 6 0 1 4 y 4 2 -4 -2 2 2 1 1 0 0 0 1 0 0 0 0 0 0 0 0 3. 0 4 2 1 4 2 0 3 1 1 3 11 (c) (4, 4) v x 2 4 5. 0 0 9. 0 0 7. T(v) (- 4 2, 0) 2 -6 - 4 - 2 -2 -4 11. 35, 7 23. (a) 1 2 3 2 (c) 2 y 3 2 1 2 (b) 1 2 3, 1 3 2 13. 0, 6, 6, 17. (a) (b) (c) 1 0 3, 15. 0, 0 (1, 2) v (3, 4) v x 2 4 1 -60 T(v) 1 (1 + 2 2 2 3 , 2- 3 2 ) x T(v) (- 3,-4) -4 19. (a) (b) (c) 1 0 2, 3 0 1 1 25. (a) 0 0 (c) 0 1 0 2 0 0 1 z (b) 3, 2, 2 (3, 2, 2) v y 4 3 x 1 2 3 4 y 1 - 3 -2 -1 x 1 2 3 T(v) (3, 2, -2) T(v) -2 (- 2, - 3) v (2, - 3) 1 27. (a) 0 0 0 1 0 0 0 1 (b) 1, 2, 1 A48 Answer Key (c) (1, -2, -1) 2 T(v) 4 x z v 1 2 3 4 y 2 35. (a) 3 2 1 0 37. (a) 1 0 39. A 0 1 0 4 2 1 3 0 1 1 0 2 0 2 1 0 0 0 1 2 1 0 1 0 0 4 ,A 5 0 0 ,A 0 1 ,A 5 x 2 y x , 0 0 1 1 (b) 9 5 1 1 1 2 1 (1, 2, -1) (b) 29. (a) 3 2 1 2 y 3 1 2 3 2 (b) 3 2 1, 3 1 2 0 7 0 1 0 5 4 2 y 2 x1 2 3 0 0 0 3 3 3 2 1 1 0 0 0 (c) 41. A ( T(v) v 3 1 - 1, 3 + 2 2 (1, 2) ( 43. A x 1 45. T x, y 1 9 10 3 10 3 10 1 10 2 3 21 7 10 , 10 31. (a) (c) (b) y 4 3 2 1 (1, 4) v x1, x2, x2 x3 47. T 1 x1, x2, x3 49. T is not invertible. 51. T is not invertible. x y 1 , 53. T x, y 5 5 x1 2x2, x2, x4, x3 x4 55. T 1 x1, x2, x3, x4 57. 9, 5, 4 61. 2, 4, 3, 3 0 0 0 1 0 0 65. 0 1 0 0 0 1 1, 5 59. 63. 9, 16, 20 0 1 0 0 0 0 67. 0 0 1 0 0 0 0 0 0 0 1 4 T(v) 1 3 5 4 5 7 ( 21 , 10 ( 10 -1 x 2 3 16 5, 13 5 33. (a) (c) 4 5 3 5 0 0 1 1 (b) (1, 4) v y 4 3 2 1 2ex 0 1 71. (a) 0 0 0 69. 3 2xex 0 0 0 0 1 0 2 1 0 3 0 0 (b) 6x x2 3 4 4x -1 -2 -3 x 3 4 5 T(v) (16 , - 13( 5 5 73. (a) True. See discussion under "Composition of Linear Transformations," pages 390391. (b) False. See Example 3, page 392. (c) False. See Theorem 6.12, page 393. Answer Key A49 1 0 0 75. 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 (b) A 1 0 1 4 4 0 9 8 3 2 3 1 3 1 3 1 3 1 6 1 3 1 3 1 3 1 3 1 2 1 1 1 2 2 1 1 77. Proof 79. Sometimes it is preferable to use a nonstandard basis. For example, some linear transformations have diagonal matrix representations relative to a nonstandard basis. 0 2 2 1 2 0 (b) v B 9. (a) 6 9 2 , Tv 1 1 3 3 4 1 3 1 2 B 4 4 Section 6.4 (page 405) 1. (a) A 4 5 3 (c) A 4 3 7 ,P 1 3 1 0 1 3 1 3 13 3 1 3 1 3 4 3 16 3 1 3 4 3 (d) 2 1 1 1 1 2 0 3 5 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 (b) A 1 2 3 11. (a) 3. (a) A (b) v 1 0 0 0 1 1 4 4 1 1 1 (c) A B (b) A 1 0 0 5. (a) A 0 1 0 1 2 1 2 1 2 1 2 1 2 1 2 1 0 0 1 0 3 1 0 , Tv B 1 0 0 2 0 ,P 0 3 2 1 2 1 1 1 0 1 0 1 0 1 1 (b) A 1 1 0 1 2 1 2 1 2 1 0 0 0 1 0 0 0 1 (d) 5 9 13. (a) 2 2 B 1 0 1 7 3 1 6 2 3 10 3 4 3 4 3 0 1 1 1 3 8 3 2 3 (b) v 3 , Tv 1 7 27 2 ,P 8 B 5 1 1 (c) A 1 5 7. (a) A 1 4 9 8 1 4 5 8 (d) 15. Proof 17. Proof 19. In A50 Answer Key 2127. Proof 29. The matrix for I relative to B and B is the square matrix whose columns are the coordinates of v1, . . . , vn relative to the standard basis. The matrix for I relative to B, or relative to B , is the identity matrix. 31. (a) True. See discussion, page 399400, and note that A P 1AP PA P 1 PP 1APP 1 A. (b) False. Unless it is a diagonal matrix; see Example 5, pages 403404. 13. (a) Vertical shear (b) y 3 2 1 (x, 2x + y) (x, y) x 1 2 3 Section 6.5 (page 413) 1. (a) (d) 3. (a) (d) 5. (a) 7. (a) (b) (b) 2, 1 (c) a, 0 c, d (e) (f) f, g (b) 3, 1 (c) 0, a g, f (e) d, c (f) (b) Reflection in the line y x Vertical contraction y 3, 5 0, b 1, 0 b, 0 y, x 15. 19. 23. 0, t : t is real t, 0 : t is real y 17. 21. 25. 1 x t, t : t is real t, 0 : t is real y 1 ( , 1( 1 2 -1 (x, y) (0, -1) (1, -1) ( , 0( 1 2 x 1 y 27. y 2 29. (-1, 2) (2, 1) (3, 1) 1 x 1 2 3 2 ( x, y2 ( x 1 9. (a) Horizontal expansion (b) y (x, y) (4x, y) 4 3 2 1 x 1 2 3 4 (-1, 0) -2 -1 -1 x 31. 1 y 33. (0, 1) (1, 1) 3 y (2, 2) 2 1 x (3, 2) 11. (a) Horizontal shear (b) y 1 1 2 3 x (x, y) (x + 3y, y) x Answer Key A51 35. (a) y 8 6 4 2 (b) y 12 10 8 6 4 2 (9, 6) (3, 2) 2 (6, 6) (6, 0) (12, 6) (7, 2) x 4 (6, 0) 57. 90 about the x-axis 59. 180 about the y-axis 61. 90 about the z-axis 0 1 0 0 0 1 63. 1 0 0 Line segment from 0, 0, 0 to 1, 1, 1 1 3 3 4 4 2 3 1 65. 0 2 2 3 1 3 4 4 2 Line segment from 0, 0, 0 to 3 3 1 4, 1 3 2, (0, 0) x (0, 0) 2 4 6 8 10 12 37. (a) y 5 4 3 2 1 (b) y 5 (6, 3) (0, 0) (2, 1) 2 4 6 4 3 2 1 (0, 3) (0, 0) 2 4 6 (12, 3) (10, 1) (12, 0) x 8 10 12 3 1 4 (12, 0) x 8 10 12 Review Exercises Chapter 6 (page 416) 1. (a) 2, 3. (a) 0, 5. Linear, 9. Not linear 4 1, 7 1 1 1 0 1 3 3 2, 2 39. T 1, 0 y 6 5 4 3 2 1 -1 2, 0 , T 0, 1 (4, 6) 0, 3 , T 2, 2 4, 6 (b) 4, 4 (b) t 3, 5 2 1 1 1 0 , T 0, 1 1, 2 0 1 1 1, 1 t, t : t is real. 1 1 2 2 7. Linear, (2, 2) 11. Linear, x 1 2 3 4 5 13. T 1, 1 15. T 0, 17. A2 19. A3 I 1 41. 43. 45. 47. Horizontal expansion Reflection in the line y x Reflection in the x-axis followed by a vertical expansion Vertical shear followed by a horizontal expansion 3 2 1 2 0 3 1 1 2 3 2 0 1 2, 0 51. 0 1 3 1 2, 1 3 2 1 2 0 3 2 0 1 0 3 2 0 1 2 cos 3 sin 3 21. (a) T: R3 R2 (c) 5 2, sin 3 cos 3 (b) 3, 2t, t : t is real 12 49. 3 53. 55. 3 2, 1, 1 23. (a) T: R2 R1 (b) 5 (c) 4 t, t : t is real 25. (a) T: R3 R3 27. (a) T: R2 R3 2, 4, 5 (b) (b) 8, 10, 4 (c) 2, 2, 2 (c) 1, 1 A52 Answer Key 29. (a) 2, 1, 0, 0 , 2, 0, 1, 2 (b) 5, 0, 4 , 0, 5, 8 31. (a) 1, 1, 1 (b) 1, 0, 1 1 33. (a) 0, 0 (b) 1, 0, 2 (c) Rank 2 (d) Nullity 3, 3, 1 35. (a) (b) 1, 0, (c) Rank 2 (d) Nullity 37. 3 39. 2 41. A 43. A 2 0 cos sin 0 ,A 1 sin cos 1 , 0, 1, 1 1 , 0, 1, 2 0 1 , 0, 1, 2 1 v: v, v0 0 69. Ker T Range R Rank 1 1 Nullity dim V 71. mn pq 73. (a) Vertical expansion (b) y (x, 2y) 1 2 0 1 1 0 ,A (x, y) cos sin sin cos 75. (a) Vertical shear (b) y x 45. T has no inverse. 47. T has no inverse. 0 0 0 0 0 0 1 0 ,A 49. A 1 1 0 1 0 51. 8 y (x, y + 3x) (- 3, 5) 6 (- 5, 3) 2 - 6 - 4 -2 -2 (0, 3) (3, 5) (5, 3) x (x, y) x (3, 0) 6 77. (a) Horizontal shear (b) y (c) Invertible (c) Invertible (x, y) (x + 2y, y) x 53. (a) One-to-one 55. (a) One-to-one 57. 0, 1, 1 3 1 , 59 A 1 1 A P 1AP 0 61. (a) 0 0 0 1 5 2 5 1 2 1 2 (b) Onto (b) Onto 1 2 1 2 1 1 3 1 1 1 1 1 79. y 81. (1, 0) 1 2 x 1 y 0 2 5 4 5 (0, 0) (3, 1) (0, 0) x (b) Answers will vary. (c) Answers will vary. -1 63. Answers will vary. 65. Proof 67. (a) Proof (b) Rank 1, nullity (c) 1 x, 1 x2, 1 x3 (0, - 1) (1, 0) -1 2 3 3 Answer Key A53 83. Reflection in the line y expansion 85. 1 87. 0 0 1, 89. 1 3 2 1 2 0 6 4 6 4 1 2 2 2 2 2 0 0 1 2 3 2 3 2, 1 4 3 4 3 2 2 2 2 2 0 2 2 2 2 0 0 0 1 0 3 2 , 1 2 3 x followed by a horizontal Chapter 7 Section 7.1 (page 432) 1. 3. 1 0 1 1 0 1 1 1 3 1 0 3 1 0 3 1 0 1 0 0 1 1 1 1 1 0 1 1 1 2 3 1 2 3 1 2 3 5 1 2 1 0 0 1 1 0 5 3 1 2 1 1 1 1 0 2c 2c c c 2 c c 1 1 1 0 , 0 0 1 1 1 0 , 1 1 1 2 0 , 0 1 1 1 , 0 0 1 1 1 1 1 1 0 1 2 1 1 2 , 0, 1 1 2 3 4 3 4 1 2 2 4 2 4 3 2 2,0 , 2 2 , ,1 , 2 2 2 5. 0 0 2 0 0 2 0 0 0 7. 0 1 9. (a) 1 1 1 (b) 1 91. 93. 0, 0, 0 , 2 2 , , 0 , 0, 2 2 2 2 , , 0 , 0, 0, 1 , 2 2 0, 2,1 , 2 2 , ,1 2 2 0 1 1 1 0 1 c c c c 3 1 3 1 , , 0, , , 95. 0, 0, 0 , 1, 0, 0 , 1, 2 2 2 2 0, 1, 1 3 , , 1, 2 2 1 2 3 1 , 2 1 3 , , 2 2 3 , 0, 1 2 3 1 , 3 2 97. (a) False. See "Elementary Matrices for Linear Transformations in the Plane," page 407. (b) True. See "Elementary Matrices for Linear Transformations in the Plane," page 407. (c) 99. (a) (b) (c) True. See discussion following Example 4, page 411. False. See "Remark," page 364. False. See Theorem 6.7, page 383. True. See discussion following Example 5, page 404. 11. (a) No (b) Yes (c) Yes (d) No 13. (a) Yes (b) No (c) Yes (d) Yes 1 2 7 0 0 15. (a) 17. (a) 4 1 0, 1, 2 (b) (b) 2 , 1, 1 1 7, 3, 1 2 , 3, 1 2 3 1 0 19. (a) 1, 1, 2, 1 ; 2, 1, 0, 0 ; 3, 0, 1, 0 (b) 2 4 1 0 21. (a) 4, 7, 4, 2 ; 2, 1, 0, 0 ; (b) 1, 1, 1, 1 3 32 0 23. (a) 3, 1, 1, 3 ; 3, 1, 0, 1 , 1, 1, 0 (b) A54 Answer Key 25. (a) (b) 27. (a) (b) 29. 35. 41. 45. 49. 2 3 6 2, 3, 2, 0 ; 6, 1, 2, 0 4 2 0 4, 5, 75. Proof 10, 2 Section 7.2 1. P 1 (page 444) 3 22 0 2, 1, 0, 0, 0 , 0, 1, 0, 0 ; 3, 0, 0, 3, 4 ; 0, 0, 0, 0, 1 2, 1 5, 5 31. 33. 37. 39. 1, 4, 4 0, 3 6 2 2 1 1 1 5 1 5 2 3 4 , P 1AP 3 4 5 1 5 2 3 1 4 1 12 1 2 1 2 5 2 1 2 1 0 2 0 0 2 0 3 5 0 0 4 0 0 0 2 0 0 3 0 0 0 3 0 0 1 4, 12 1 1 2, 3 3. P 1 , P 1AP 0, 0, 0, 21 1 0 , P 1AP 0 15 27 5. P 1 8 2 43. 2 3 7 2 Exercise Trace of A 15 7 1 17 0 4 19 6 6 21 7 8 27 23 3 48 25 8 27 7 0 51. Proof 53. Assume that is an eigenvalue of A, with corresponding eigenvector x. Because A is invertible (from Exercise 0. Then, Ax x implies that x A 1Ax 52), 1 x 1x, A A which in turn implies that 1 x A 1x. So, x is an eigenvector of A 1, and its corresponding eigenvalue is 1 . 55. Proof 57. Proof 59. a 0, d 1 or a 1, d 0 61. (a) False. (b) True. See discussion before Theorem 7.1, page 424. (c) True. See Theorem 7.2, page 426. 63. Dim 3 65. Dim 1 d x e ex 1 ex 67. T e x dx 2, 3 2x; 4, 5 10x 2x2; 69. 6, 1 2x 71. 0, 1 1 0 1 , 0 0 1 ; 1 3, 1 2 0 0 5 47. Determinant of A 0 0 1 3 7. P 9. 11. 13. 1 1 0 , P 1AP 15. 17. 19. 21. 23. 1 0 0 0, and the dimension There is only one eigenvalue, of its eigenspace is 1. The matrix is not diagonalizable. 1, and the dimension There is only one eigenvalue, of its eigenspace is 1. The matrix is not diagonalizable. There are two eigenvalues, 1 and 2. The dimension of the eigenspace for the repeated eigenvalue 1 is 1. The matrix is not diagonalizable. There are two repeated eigenvalues, 0 and 3. The eigenspace associated with 3 is of dimension 1. The matrix is not diagonalizable. 0, 2 The matrix is diagonalizable. 0, 2 Insufficient number of eigenvalues to guarantee diagonalizability 1 3 P (The answer is not unique.) 2 1 3 1 P (The answer is not unique.) 1 1 7 4 2 1 1 3 1 0 1 1 0 0 1 1 0 1 1 1 (The answer is not unique.) 25. P 27. P (The answer is not unique.) 73. The only possible eigenvalue is 0. Answer Key A55 29. P 3 2 0 4 4 4 1 1, 1 2 0 0 0 3 1 5 10 2 0 2 1 1 (The answer is not unique.) 25. P 3 3 6 3 2 3 1 3 2 3 1 3 2 3 2 3 2 3 2 3 1 3 6 3 3 3 (The answer is not unique.) 0 0 (The answer is not unique.) 0 1 37. 41. 1 188 126 x ,x 378 253 31. A is not diagonalizable. 27. P (The answer is not unique.) 33. P 35. 1 , 1, 1 29. 3 3 3 3 3 3 2 2 2 2 0 6 6 6 6 6 3 39. (a) and (b) Proof 43. 384 384 128 256 512 256 384 1152 640 (The answer is not unique.) 2 2 2 2 0 0 0 0 2 2 2 2 2 2 2 2 0 0 0 0 2 2 2 2 31. P 45. (a) True. See the proof of Theorem 7.4, pages 436437. (b) False. See Theorem 7.6, page 442. 47. Yes, the order of elements on the main diagonal may change. 4955. Proof k is 0, 0 . By 57. The eigenvector for the eigenvalue Theorem 7.5, the matrix is not diagonalizable because it does not have two linearly independent vectors. (The answer is not unique.) 33. (a) True. See Theorem 7.10, page 453. (b) True. See Theorem 7.9, page 452. 35. Proof 37. Proof 1 cos sin 39. A 1 cos2 sin2 sin cos cos sin AT sin cos 41. Proof Section 7.3 1. Symmetric 5. Symmetric 9. 13. (page 456) 3. Not symmetric 2, dim 1 7. 4, dim 1 11. 2, dim 4, dim 2 1 Section 7.4 1. x2 (page 472) 2, dim 2 3, dim 1 1, dim 1 1 2, dim 1 2, dim 3. x2 1 1 17. Not orthogonal 21. Orthogonal 7. x 15. Orthogonal 19. Orthogonal 23. P 20 , x3 5 84 12 , x3 6 8 t 4 1 900 60 , x3 50 10 10 60 84 6 5. x t 2 1 960 90 , x3 30 2340 720 45 2 9. x2 2200 540 30 2 2 2 2 2 2 2 2 (The answer is not unique.) 11. x2 A56 Answer Key 13. y1 y2 19. y1 y2 23. y1 y2 y3 25. y1 y2 y3 27. y1 y2 31. 1 0 C1e2t C2et C1et C2e2t 3C1e 2C1e C1et 15. y1 y2 y3 4C2e2t 2t 2t C1e t C2e6t C3et 21. y1 y2 C3e6t 2C3e6t 17. y1 y2 y3 C1e C1e t t C1e2t C2e t C3et C2e3t C2e3t 53. A x 2 1 0 0 y 0 2 1 2 0 1 , 2 3z 2 1 0 5C2e4t 10C2e4t 2C2e4t 55. Let P 2C2e2t C2e2t y2 y2 33. 9 5 7C3e3t 8C3e3t 2C3e3t 29. y1 y2 y3 5 4 y2 y3 4y2 0 35. 5 5 10 y1 0 1 2 37. A 3 2 2 3 2 5 , 2 1 10 3 10 16 12 , 39. A 13 3 3 , 7 3 3 a b be a 2 2 orthogonal matrix such c d 1. Define 0, 2 as follows. that P 0. (i) If a 1, then c 0, b 0, and d 1, so let 1, then c 0, b 0, and d 1, so let (ii) If a . 2. (iii) If a 0 and c > 0, let arccos a , 0 < 2 (iv) If a 0 and c < 0, let arccos a , < 2 . 3 2 (v) If a 0 and c > 0, let arccos a , 2 < . 2 (vi) If a 0 and c < 0, let arccos a , < 3 2. In each of these cases, confirm that a b cos sin . P c d sin cos (page 474) 1 2 5 , 2 3 10 1 10 12 , 9 1 1 4, 2 16, 3 2 1 2 3 5 4 5 4 5 3 5 Review Exercises Chapter 7 P P 1 2 3 2 25, P 1. (a) 2 9 0 (b) 3, 3 3 is 1, 5 and a basis for (c) A basis for 3 is 1, 1 . 4 82 0 4, 8 3. (a) (b) 4 is 1, 2, 1 and a basis for (c) A basis for 8 is 4, 1, 0 , 3, 0, 1 . 2 3 1 0 5. (a) 1, 2, 3 (b) 1 is 1, 2, 1 , a basis for 2 (c) A basis for 3 is 0, 1, 0 . is 1, 0, 0 , and a basis for 12 32 0 1, 3 7. (a) (b) 1 is 1, 1, 0, 0 , 0, 0, 1, 1 (c) A basis for 3 is 1, 1, 0, 0 , 0, 0, 1, 1 . and a basis for 9. Not diagonalizable 1 0 1 0 1 0 (The answer is not unique.) 11. P 1 0 1 41. A 43. 45. 47. 49. 0, 2 Ellipse, 5 x 2 15 y 2 45 0 Hyperbola, 25 x 2 15 y 2 50 8y 4 0 Parabola, 4 y 2 4x 1 2 0 Hyperbola, x 2 y 2 3 2x 0 0 , 8 8z 2 2y 6 0 51. A 2x 2 3 1 1 3 0 0 4 y 2 16 0 Answer Key A57 13. The characteristic equation of A cos sin sin cos 51. P 15. 17. 19. 21. 23. 25. 29. 2 cos 1 0. The roots of this equation is 2 are cos cos2 1. If 0 < < , then 1 < cos < 1, which implies that cos 2 1 is imaginary. A has only one eigenvalue, 0, and the dimension of its eigenspace is 1. So, the matrix is not diagonalizable. A has only one eigenvalue, 3, and the dimension of its eigenspace is 2. So, the matrix is not diagonalizable. 0 1 P 1 0 1 of Because the eigenspace corresponding to matrix A has dimension 1, while that of matrix B has dimension 2, the matrices are not similar. Both orthogonal and symmetric Symmetric 27. Neither 1 2 5 5 P (The answer is not unique.) 1 2 5 5 1 2 0 1 0 1 2 0 1 2 1 2 1 2 1 2 1 2 53. (a) a b c 0 (b) Dim 1 if a 0, b 0, c 0. Dim 2 if exactly one is 0. Dim 3 if exactly two are 0. 55. (a) True. See "Definitions of Eigenvalue and Eigenvector," page 422. (b) False. See Theorem 7.4, page 436. (c) True. See "Definition of a Diagonalizable Matrix," page 435. 100 25 2 , x3 , x t 57. x2 25 25 1 59. x2 4500 300 , x3 50 1440 108 , x3 90 2C1 C1 C2et 1500 4500 , x 50 6588 1296 81 65. y1 y2 y3 C1et C1et C3 y 24 t 12 1 61. x2 63. y1 y2 C2e C2e t t 31. P 0 1 2 1 67. A 37. 1 1 1 4, 2, 4 (The answer is not unique.) 33. 39. 3 2 5, 5 4 5 7 16 , 16 , 16 3 2 1 2 35. 3 2 5, 5 3 2 1 2 1 2 2 41. Proof 1 9 4 43. A 45. A2 0 0 , 1 0, 2 9 4 P 1 2 1 2 y 2 -2 -2 x' x 2 y' 56 40 , A3 20 4 47. (a) and (b) Proof 368 304 152 88 49. Proof 5x 6 A58 Answer Key 0 69. A 1 2 1 2 1 2 2 1 2 3 y 12. 0 1 2 1 2 2 2 1 x 2 3 1 1 (three times), 0 0 3 1 1 1 (The answer is not unique.) 13. P 14. P 0, 1, 0 , 1, 1, 1 , 2, 2, 3 1 2 1 2 1 3 1 3 1 3 C1et C2e3t 4 4 4 4 1800 120 , x3 60 6300 1440 48 1 2 1 2 1 2 0 1 2 1 6 2 6 1 6 x' y' x y 4 (page 479) 15. Cumulative Test Chapters 6 and 7 1. 2. 3. 4. 5. Yes, T is a linear transformation. No, T is not a linear transformation. (a) 1, 1, 0 (b) 5, t s, s, t, t : s, t are real (a) Span 0, 1, 0, 1 , 1, 0, (b) Span 1, 0 , 0, 1 (c) Rank 2, nullity 2 1 1 0 1 2 1 2 1 16. 1, 0 17. y1 y2 18. 1 6. 0 1 7. 8. T 9. 0 1 1 1 2 1 2 19. x2 , T 1, 1 1 3x 1 3 y, 0, 0 , T 2 3x 1 3y 2, 2 2, 2 20. x, y 2 1 , T 0, 1 1 1 2 10. (a) A 1 4 7 15 (c) A 6 12 1 , T (e) v B 1 1 1, 0 ; 0, 11. 0 1 0 2 1, 0, 1 1 1 9 6 3 3 2, 1 1 1 1 2 (b) P (d) v B is an eigenvalue of A if there exists a nonzero vector x such that Ax x. x is called an eigenvector of A. If A is an n n matrix, then A can have n eigenvalues, possibly complex and possibly repeated. 21. P is orthogonal if P 1 PT. The possible eigenvalues of the determinant of an orthogonal matrix are 1 and 1. 2226. Proof 27. 0 is the only eigenvalue. 1 1 ; 3 ...
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