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6
ISM: Chapter Linear Algebra
Chapter 6 6.1
1. Fails to be invertible; since det 2. Invertible; since det 3. Invertible; since det 1 2 = 6  6 = 0. 3 6
2 3 = 10  12 = 2. 4 5 3 5 = 33  35 = 2. 7 11 1 4 = 8  8 = 0. 2 8
4. Fails to be invertible; since det 2 5. Invertible; since det 0 0 6 6. Invertible; since det 5 3
7. This matrix is clearly not invertible, so the determinant must be zero.
5 7 11 7 = 2 11 5 + 0 + 0  0  0  0 = 110. 0 5 0 0 4 0 = 6 4 1 + 0 + 0  0  0  0 = 24. 2 1
8. This matrix fails to be invertible, since the det(A) = 0. 0 1 2 9. Invertible; since det 7 8 3 = 0 + 3 6 + 2 7 5  7 4  2 8 6 = 36. 6 5 4 1 1 1 10. Invertible; since det 1 2 3 = 1 2 6 + 1 3 1 + 1 1 3  3 3 1  2 1 1  6 1 1 = 1. 1 3 6 11. det 12. det k 3 2 3 = 0 when 4k = 6, or k = 2 . 4
k = 0 when k 2 = 4, or k = 2, 2. 4 k 3 5 13. det 0 2 6 = 8k, so k = 0 will ensure that this matrix is invertible. 0 0 4 290
1 k
ISM: Linear Algebra
Section 6.1
14.
15.
16.
17.
18.
19.
20.
21.
22.
4 0 0 det 3 k 0 = 0, so the matrix will never be invertible, no matter which k is chosen. 2 1 0 0 k 1 1 det 2 3 4 = 6k  3. This matrix is invertible when k = 2 . 5 6 7 1 2 3 det 4 k 5 = 60 + 84 + 8k  18k  35  64 = 45  10k. So this matrix is invertible 6 7 8 when k = 4.5. 1 1 1 det 1 k 1 = 2k 2  2 = 2(k 2  1) = 2(k  1)(k + 1). So k cannot be 1 or 1. 1 k2 1 0 1 k 5 det 3 2k 5 = 30 + 21k  18k 2 = 3(k  2)(6k + 5). So k cannot be 2 or  6 . 9 7 5 1 1 k det 1 k k = k 3 + 2k 2  k = k(k  1)2 . So k cannot be 0 or 1. k k k 1 k 1 det 1 k + 1 k + 2 = (k + 1)(2k + 4) + k(k + 2) + (k + 2)  (k + 1)  k(2k + 4)  1 k + 2 2k + 4 (k + 2)(k + 2) = (k + 1)(3k + 6)  (3k 2 + 9k + 5) = 1. Thus, A will always be invertible, no matter the value of k, meaning that k can have any value. k 1 1 det 1 k 1 = k 3  3k + 2 = (k  1)2 (k + 2). So k cannot be 2 or 1. 1 1 k cos k 1  sin k 2 0 = 2 cos2 k + 2 sin2 k = 2. So k can have any value. det 0 sin k 0 cos k 1 0 2 = (1  )(4  ) = 0 if is 1 or 4. 4 0 = (2  )() = 0 if is 2 or 0. 0 291
23. det(A  I2 ) = det 24. det(A  I2 ) = det
2 1
Chapter 6 4 4 4 2
ISM: Linear Algebra 2 = (4  )(6  )  8 = (  8)(  2) = 0 if is 2 or 6 2 = (4  )(7  )  4 = (  8)(  3) = 0 if is 3 or 7
25. det(A  I2 ) = det 8. 26. det(A  I2 ) = det 8.
27. A  I3 is a lower triangular matrix with the diagonal entries (2  ), (3  ) and (4  ). Now, det(A  I3 ) = (2  )(3  )(4  ) = 0 if is 2, 3 or 4. 28. AI3 is an upper triangular matrix with the diagonal entries (2), (3) and (5). Now, det(A  I3 ) = (2  )(3  )(5  ) = 0 if is 2, 3 or 5. 3 5 6 29. det(A  I3 ) = det 0 4 2 = (3  )(  8)(  3) = 0 if is 3 or 8. 0 2 7 4 2 0 6 0 = (4  )(6  )(3  )  8(3  ) 30. det(A  I3 ) = det 4 5 2 3 = (3  )(8  )(2  ) = 0 if is 3, 8 or 2. 31. This matrix is upper triangular, so the determinant is the product of the diagonal entries, which is 24. 32. This matrix is upper triangular, so the determinant is the product of the diagonal entries, which is 210. 33. By Fact 6.1.8, the determinant is equal to det 1 2 2 3 det = (716)(1021) = 99. 8 7 7 5 4 5 1 4 det 3 6 2 3 = (24  15)(3  8) =
34. By Fact 6.1.8, the determinant is equal to det 45. 35. We will use 2 2 det 6 7 18.
Fact 6.1.5 and expand down the third column. Our determinant equals: 3 2 0 3. 0 4
Then we expand down the second column, so we have 2(3) det
6 3 = 6(24  21) = 7 4
36. We use Fact 6.1.5, first expanding down the second column. Our determinant equals: 292
ISM: Linear Algebra
Section 6.1
2 2 2 2 det 1 2 2 = 2(8 + 4 + 2  4  4  4) = 4. 1 1 2 5 6 7 5 4 37. Here we use Fact 6.1.8 to find det det 0 1 2 = (35  24)(5) = 55. 6 7 0 0 1 1 2 3 6 5 38. By Fact 6.1.8, the determinant is equal to det 3 0 4 det 5 6 2 1 2 = (9 + 16  12  4)(36  25) = 99. 0 1 = 120. 3 0 39. We repeatedly expand down the first column, finding the determinant to be 5 4(2) det
40. We repeatedly expand down the first column, finding the determinant to be 5(4)(3) det 0 2 = 120. 1 0 0 3 0 0 1 5 4 3 2 7 , then down the second 6 4
0 1 41. Here we expand down the fourth column: 2 det 2 0 0 1 2 1 2 column: 2(3) det 2 4 6 = 2(3)(2) det 3 4 0 3 4 42. We first expand 1 1 0 0 2 0 3 det 2 3 4 3 4 5 1 0 3(2) det 2 4 3 5
= 12(4  6) = 24.
43. For a 2 2 matrix A =
a b a b = ad  bc = det(A). , we have det(A) = det c d c d Use Sarrus' rule to see that det(A) =  det(A) for a 3 3 matrix. We may conjecture 293
across the fourth row to obtain 1 0 , then across the second row to obtain 0 6 1 0 = 6(24 + 10  12) = 132. 6
Chapter 6
ISM: Linear Algebra
that det(A) = det(A) for an n n matrix with even n, and det(A) =  det(A) when n is odd; in both cases we can write det(A) = (1)n det(A). A proof of this conjecture can be based upon the following rule: If a square matrix B is obtained from matrix A by multiplying all the entries in the ith row of A by a scalar k, then det(B) = k det(A). We can show this by expansion along the ith row:
n n
det(B) =
j=1
(1)i+j bij det(Bij ) =
j=1 n
(1)i+j kaij det(Aij )
=k
j=1
(1)i+j aij det(Aij ) = k det(A).
We can obtain A by multiplying all the entries in the first row of A by k = 1, then all the entries in the second row, and so forth down to the nth row; each time the determinant gets multiplied by k = 1. Thus, det(A) = k n det(A) = (1)n det(A), as claimed. 44. det(kA) = k n det(A) The argument is analogous to the one in Exercise 43. 45. If A = a b a , then det(AT ) = det c d b T det(A ) = det(A). a1 a3 c d = ad  cb = det(A). It turns out that a4 a3
2
46. Let A =
a2 . If a1 a4  a2 a3 = 0, then A1 = a4
1 det(A) 2
1 det(A)
a2 . a1 det(A) so det(A1 ) =
By Exercise 44, det(A1 ) =
1 det(A) .
(a1 a4  a2 a3 ) =
1 det(A)
47. We have det(A) = (ah  cf )k + bef + cdg  aeg  bdh. Thus matrix A is invertible for all k if (and only if) the coefficient (ah  cf ) of k is 0, while the sum bef + cdg  aeg  bdh is nonzero. A numerical example is a = c = d = f = h = g = 1 and b = e = 2, but there are infinitely many other solutions as well. 48. Consider A = 1 0 ,B = 0 0 0 0 ,C = 1 0 0 1 ,D = 0 0 0 0 0 1 so det(A) = det(B) = A C B D = 1.
det(C) = det(D) = 0 hence det(A) det(D)  det(B) det(C) = 0 but det
49. The kernel of T consists of all vectors x such that the matrix [x v w] fails to be invertible. This is the case if x is a linear combination of v and w, as discussed on Pages 247 and 248. Thus ker(T ) = span(v, w). The image of T isn't {0}, since T (v w) = 0, for example. Being a subspace of R, the image must be all of R. 294
ISM: Linear Algebra
Section 6.1
50. Fact 6.1.1 tells us that det[u v w] = u(vw) = u cos() vw = u cos() v sin() w = cos() sin(), where is the angle enclosed by vectors u and v w, and is the angle between v and w. Thus det[u v w] can be any number on the closed interval [1, 1]. 51. We expand down the first column: 2 1 det(Mn ) = 1 det 0 . . . 1 2 1 .. . 0 1 2 .. . 0 An2 1 0 0 1 2 0 0  1 det 0 1 . . . .. . .. . . . . 0 1 2 0 0 0 1 2 .. . 0 0 0 0 . .. . . . 1 2
0
0 this equals n  (n  1) = 1. 2 0 1 0 52. det(Dn ) = det 0 . . . 0 1 0 0 1 0 0 2 1 0 1 0 . . . An3 0 0 0 0 0 0 2 1 0 0 . . .
1 0 1 0 = det(An1 )  det . . . 0
= det(An1 )  1 det(An2 ). Using Example 9, 0 0 0 0 0 0
Then we expand down the first column: 2 1 0 = 2 det 0 . . . 0 1 0 0 2 1 0 1 0 . . . An3 0 0 0 0 0 0 2 0 0 0 + det 0 . . . 0 1 0 0 1 0 0 1 0 . . . An3 0 0 0 0
0 0 0
= 2 det(An1 )  2 det(An3 ) = 2(n)  2(n  2) = 4. det(Dn ) will never be 0. 295
Chapter 6 2 0 0 0 0 53. det(En ) = det 1 . . . 0 2 1 0 = 2 det(An1 )  1 det 0 . . . 0 0 1 0 1 0 1 1 0 .. . 0 0 1 0 1 1 0 .. . 0 0 0 0 0 2 1 .. . 0 0 0 0 0 0 0 0 0 0 0 1 2 .. . 0 0 .. . .. . .. . .. . .. . .. . .. . An4 0 0 0 0 0 .. . 2 1 0
ISM: Linear Algebra
An1
0 1 2 0 0 .. . 0 0 0 1 2 0 0 .. . 0 0
0
0 0 1 = 2 det(An1 )  det 2 1 0 det(An4 ) (by Fact 6.1.8) = 2 det(An1 )  3 det(An4 ) 1 2 1 = 2n  3(n  3) = 2n  3n + 9 = 9  n. So, det(E9 ) = 0.
0 2 1 0 = 2 det(An1 )  1 det 0 . . . 0 0
0 0 0 0 0 0
2
0 0 0 0 . . . 1
54. a. We will expand down the first column:
6 0 0 1 0 dn = det(Mn ) = 5 det(Mn1 )  1 det . . . 0 0 296
Mn2
0 0
ISM: Linear Algebra = 5 det(Mn1 )  1(6) det(Mn2 ). b. d1 = det[2] = 2, d2 = det 5 6 = 4, 1 2
Section 6.1
d3 = 5(4)  6(2) = 8, d4 = 5(8)  6(4) = 16. c. dn = 2n . The base case has already been shown. If we assume that dn1 = 2n1 and dn2 = 2n2 , then dn = 5dn1  6dn2 = 5(2n1 )  6(2n2 ) = 5(2n1 )  3(2n1 ) = 2(2n1 ) = 2n . 1 1 0 55. a. dn = det 0 0 . . . 0 1 1 0 = 1 det 0 . . . 0 1 0 1 1 1 1 0 1 0 0 . . . . . . 0 0 1 0 1 1 1 1 0 1 . . . . . . 0 0 0 0 1 1 1 .. . 0 0 0 1 1 .. . 0 0 0 0 1 1 .. . 0 .. . .. . 0 0 0 .. . 0 .. . 0 . .. . . . 1
1 0 0 0 1 0 = dn1  1 det 0 Mn2 . . . 0 b. d1 = det[1] = 1. d2 = det
1 0 1 0 0 0  1 det 0 0 . . . . . . 0 1
0 0 1 1 1 1 0 1 . . . . . . 0 0
0 0 1 1 .. . 0
.. . .. .
1
0 = dn1  dn2
0 0 0 0 . . .
1 1 = 0. d3 = 0  1 = 1. d4 = 1  0 = 1. 1 1 d5 = 1  (1) = 0. d6 = 0  (1) = 1. d7 = 1  0 = 1. d8 = 1  1 = 0. . . .. Note that d7 = d1 = 1 and d8 = d2 = 0, so that the values are repeated, with a period of 6, meaning that dk = dk+6 for all k.
c. Following part b, we have d100 = d10016(6) = d4 = 1. 297
Chapter 6 0 . . 56. a. dn = det . 0 1
ISM: Linear Algebra
Mn1
Now, if we expand down the first column, we find that dn = det(Mn1 ) if n is odd, or dn =  det(Mn1 ) if n is even. We can model this by simply saying: dn = (1)n1 dn1 .
.
b. d1 = 1, d2 = 1, d3 = 1, d4 = 1, d5 = 1, d6 = 1, d7 = 1, d8 = 1. We notice the pattern that it keeps switching between 1 and 1 with every other increase. We see that dn+4 = dn . c. By the periodicity in part b, we see that d100 will be equal to d10024(4) = d4 = 1.
57. Repeatedly expanding down the first column, we see that the determinant will be 1 or 1, since it is the product of n terms that are all 1 or 1.
58. a. If a, b, c, d are distinct prime numbers, then ad = bc, since the prime factorization of a b = 0: No matrix of the required form a positive integer is unique. Thus det c d exists. b. We are looking for a noninvertible matrix A = [u v w] whose entries are nine distinct prime numbers. The last column vector, w, must be redundant; to keep things simple, we will make w = u + 2v. Now we have to pick six distinct prime entries for the first two columns, u and v, such that the entries of w = u 2v are prime as well. This can + 7 2 11 be done in many different ways; one solution is A = 17 3 23 . 19 5 29 59. Let Mn be the number of multiplications required to compute the determinant of an nn matrix by Laplace expansion. We will use induction on n to prove that Mn > n!, for n 3. In the lowest applicable case, n = 3, we can check that M3 = 9 and 3! = 6. Now let's do the induction step. If A is an n n matrix, then det(A) = a11 det(A11 ) + + (1)n+1 an1 det(An1 ), by Definition 6.1.4. We need to compute n determinants of (n  1) (n  1) matrices, and then do n extra multiplications ai1 det(Ai1 ), so that Mn = nMn1 + n. If n > 3, then Mn1 > (n  1)!, by induction hypothesis, so that Mn > n(n  1)! + n > n!, as claimed. 298
ISM: Linear Algebra
Section 6.2
60. To compute det(A) for an n n matrix A by Laplace expansion, det(A) = a 11 det(A11 )  a21 det(A21 ) + + (1)n+1 an1 det(An1 ), we first need to compute the n minors, which requires nLn1 operations; then we compute the n products ai1 det(Ai1 ); and finally we have to do n  1 additions. Altogether, Ln = nLn1 + n + (n  1) = nLn1 + 2n  1. Now we can prove the formula
Ln n! 1 1 1 1 = 1 + 1 + 2! + 3! + + (n1)!  n! by induction on n. 1 For n = 2, the formula gives L2 = 1 + 1  2 , or L2 = 3, which is correct: We have to 2 perform 2 multiplications and 1 addition to compute the determinant of a 2 2 matrix.
For an n n matrix A we can use the recursive formula derived above to see that nLn1 +2n1 Ln1 Ln 2 1 = (n1)! + (n1)!  n! . Applying the induction hypothesis to the n! = n!
Ln1 2 1 1 first summand, we find that Ln = (n1)! + (n1)!  n! = 1 + 1 + 2! + n! 1 2 1 1 1 1 1 (n1)! + (n1)!  n! = 1 + 1 + 2! + 3! + + (n1)!  n! , as claimed. 1 3!
++
1 (n2)!

1 1 Now recall from the theory of Taylor series in calculus that e = e1 = 1 + 1 + 2! + 3! + 1 1 1 1 1 1 + (n1)! + n! + . Thus Ln = (1 + 1 + 2! + 3! + + (n1)!  n! )n! < en!, as claimed.
6.2
1. 1 1 1 1 A = 1 3 3 I B = 0 2 2 5 2I 0 rithm 6.2.3b. 1 2 3 1 A = 1 6 8 I B = 0 2 4 0 +2I 0 rithm 6.2.3b. 1 1 3 2 4 0 1 6 4 8 I B= A= I 0 1 3 0 0 0 2 6 4 12 2I Algorithm 6.2.3b. 1 1 2 2 1 6 +I 1 2 A= 2 1 14 10 2I 2 6 10 33 +2I 1 1 2 2 3 4 0 1 0 3 10 14 3II 0 4 14 29 4II 1 1 2 2 . Now det(A) = det(B) = 6, by Algo0 3 2 3 4 5 . Now det(A) = det(B) = 24, by Algo0 6 3 2 4 3 2 4 . Now det(A) = det(B) = 24, by 0 2 4 0 0 4
2.
3.
4.
299
Chapter 6 1 1 2 1 1 2 2 0 1 3 0 1 3 4 B = 0 0 1 0 0 1 2 0 0 0 0 0 2 13 2III Algorithm 6.2.3b. 1 0 5. After three row swaps, we end up with B = 0 0 det(A) = (1)3 det(B) = 24. 1 1 1 1 1 1 4 4 I 6. A = 1 1 2 2 I 1 1 8 8 I
ISM: Linear Algebra 2 4 . Now det(A) = det(B) = 9, by 2 9 2 2 0 0 3 3 3 0 4 4 . Now, by Algorithm 6.2.3b, 4 4
1 1 1 1 0 2 1 3  2 0 0 3 3 0 2 7 9 1 3 9 1 3 6
3 2 3 2
1 1 1 1 0 0 3 3 swap: 0 2 1 3 II III 0 2 7 9
1 1 1 0 1 1 2 0 0 3 0 2 7 1 1 1 1 2 0 3 0 6
+2II 1 0 B= 0 0 1 1 1 1 2 0 3 0 0 1 3 12
3 2
det(A) =  1 (1) det(B) = 72, by Algorithm 6.2.3b. 2
1 0 0 0
2III
7. After two row swaps, we end up with an upper triangular matrix B with all 1's along the diagonal. Now det(A) = (1)2 det(B) = 1, by Algorithm 6.2.3b. 8. After four row swaps, we end up with an upper triangular matrix B with all 1's along the diagonal, except for a 2 in the bottom right corner. Now det(A) = (1)4 det(B) = 2, by Algorithm 6.2.3b. 300
ISM: Linear Algebra
Section 6.2
9. If we subtract the first row from every other row, then we have an upper triangular matrix B, with diagonal entries 1, 1, 2, 3 and 4. Then det(A) = det(B) = 24 by Algorithm 6.2.3b. 1 1 10. A = 1 1 1 1 0 0 0 0 1 1 1 1 2 3 4 5 I 3 6 10 15 I 4 10 20 35 I 5 15 35 70 I
1 1 1 1 1 2 3 4 2 5 9 14 2II 3 9 19 34 3II 4 14 34 69 4II
1 0 0 0 0
1 1 0 0 0 1 1 0 0 0
1 1 1 2 3 4 1 3 6 3 10 22 3III 6 22 53 6III
1 0 0 0 0
1 2 1 0 0
1 1 1 1 3 4 0 1 3 6 B = 0 0 1 4 0 0 4 17 4IV 0 0
1 2 1 0 0
1 3 3 1 0
1 4 6 . 4 1
Now det(A) = det(B) = 1 by Algorithm 6.2.3b. 11. By Fact 6.2.1a, the desired determinant is (9)(8) = 72. 12. By Fact 6.2.1b, the desired determinant is 8. 13. By Fact 6.2.1b, applied twice, since there are two row swaps, the desired determinant is (1)(1)(8) = 8. 14. By Fact 6.2.1c, the desired determinant is 8. 15. By Fact 6.2.1c, the desired determinant is 8. 16. This determinant is 0, since the first row 2 3 17. The standard matrix of T is A = 0 2 0 0 is twice the last. 0 6 , so that det(T ) = det(A) = 8. 2
301
Chapter 6 1 is A = 0 0 1 is A = 0 0 1 0 is M = 0 0 1 0 is A = 0 0
ISM: Linear Algebra 2 4 3 12 , so that det(T ) = det(A) = 27. 0 9 0 0 1 0 , so that det(T ) = det(A) = 1. 0 1 0 0 0 0 1 0 , so that det(L) = det(M ) = 1. 1 0 0 0 0 1 0 0 0 1 0 0 , so that det(T ) = det(A) = 1. 0 1 0 0 0 1
18. The standard matrix of T
19. The standard matrix of T
20. The standard matrix of L
21. The standard matrix of T
22. Using Exercises 19 and 21 as a guide, we observe that the standard matrix A of T is diagonal, of size (n + 1) (n + 1), with diagonal entries (1)0 , (1)1 , (1)2 , . . . , (1)n . Thus det(T ) = det(A) = (1)1+2++n = (1)n(n+1)/2 . 23. Consider the matrix M of T with respect to a basis consisting of n(n + 1)/2 symmetric matrices and n(n  1)/2 skewsymmetric matrices (see Exercises 54 and 55 or Section 5.3). Matrix M will be diagonal, with n(n + 1)/2 entries 1 and n(n  1)/2 entries 1 on the diagonal. Thus, det(T ) = det(M ) = (1)n(n1)/2 . 24. The standard matrix of T is A = 2 3 , so that det(T ) = det(A) = 13. 3 2 2 0 0 25. The standard matrix of T is A = 0 2 3 , so that det(T ) = det(A) = 16. 0 0 4 2 4 0 1 0 0 1 0 0 26. The matrix of T with respect to the basis , , is A = 2 4 2 , 0 0 1 0 0 1 0 4 6 so that det(T ) = det(A) = 16. 27. The matrix of T with respect to the basis cos(x), sin(x) is A = det(T ) = det(A) = a2 + b2 . 3 2 28. The matrix of T with respect to the basis 1 , 0 is A = 1 0 det(T ) = det(A) = 14. 302 b a , so that a b 6 10 , so that 5 6
ISM: Linear Algebra
Section 6.2
29. Expand down the first column, realizing that all but the first contribution are zero, since a21 = 0 and Ai1 has two equal rows for all i > 2. Therefore, det(Pn ) = det(Pn1 ). Since det(P1 ) = 1, we can conclude that det(Pn ) = 1, for all n.
1 1 1 30. a. f (t) = det a b t = (ab2  a2 b) + (a2  b2 )t + (b  a)t2 so f (t) is a quadratic a 2 b 2 t2 function of t. The coefficient of t2 is (b  a). b. In the cases t = a and t = b the matrix has two identical columns. It follows that f (t) = k(t  a)(t  b) with k = coefficient of t2 = (b  a). c. The matrix is invertible for the values of t for which f (t) = 0, i.e., for t = a, t = b. 31. a. If n = 1, then A = 1 a0 1 , so det(A) = a1  a0 (and the product formula holds). a1
b. Expanding the given determinant down the rightmost column, we see that the coefficient k of tn is the n  1 Vandermonde determinant which we assume is
n1i>j
(ai  aj ).
Now f (a0 ) = f (a1 ) = = f (an1 ) = 0, since in each case the given matrix has two identical columns, hence its determinant equals zero. Therefore and f (t) =
n1i>j
(ai  aj ) (t  a0 )(t  a1 ) (t  an1 ) (ai  aj ),
det(A) = f (an ) =
ni>j
as required.
32. By Exercise 31, we need to compute
i>j
(ai aj ) where a0 = 1, a1 = 2, a2 = 3, a3 = 4, a4 =
5 so
i>j
(ai  aj ) = (2  1)(3  1)(3  2)(4  1)(4  2)(4  3)(5  1)(5  2)(5  3)(5  4) = 288. 303
Chapter 6 1 ai 2 column of the given matrix as ai ai . . .
ISM: Linear Algebra
33. Think of the ith
n1 n1 n1 a1 a2 an is a Vandermonde determinant (see Exercise 31), and we get n
1 a1 2 nant can be written as (a1 a2 an ) det a1 . . . ai
i=1
n1 ai 1 a2 a2 2 . . .
, so by Fact 6.2.1a, the determi 1 an a2 n . . . . The new determinant
i>j
(ai  aj ).
34. a. The hint pretty much gives it away. Since the columns of matrix kernel of [ In M ], we have [ In M] B In
B In
are in the
= B  M = 0, and M = B, as claimed.
. . b. If B = A1 we get rref[A. n ] = [In . 1 ] which tells us how to compute A1 (see Fact .I .A 2.3.5).
35.
x1 x2
(a1 b2  a2 b1 )  x1 (b2  a2 ) + x2 (b1  a1 ) = 0. We can see that
1 must satisfy det x1 x2
1 a1 a2
1 b1 = 0, i.e., must satisfy the linear equation b2
x1 a1 x1 b = and = 1 x2 a2 x2 b2 has two identical columns in these cases.
satisfy this equation, since the matrix
36. Expanding down the first column we see that the equation has the form A  Bx1 + Cx2  D(x2 + x2 ) = 0. If D = 0 this equation defines a circle; otherwise it 1 2 is a line. From Exercise 35 we know that D = 0 if and only if the three given points c b a1 , 1 , 1 are collinear. Note that the circle or line runs through the three given c2 b2 a2 points. 304
ISM: Linear Algebra
Section 6.2
37. Applying Fact 6.2.4 to the equation AA1 = In we see that det(A) det(A1 ) = 1. The only way the product of the two integers det(A) and det(A1 ) can be 1 is that they are both 1 or both 1. Therefore, det(A) = 1 or det(A) = 1. 38. det(AT A) = det(AT ) det(A) = [det(A)]2 = 9 Fact 6.2.4 Fact 6.2.7
39. det(AT A) = det(AT ) det(A) = [det(A)]2 > 0 Fact 6.2.4 Fact 6.2.7
40. By Exercise 38, det(AT A) = [det(A)]2 . Since A is orthogonal, AT A = In so that 1 = det(In ) = det (AT A) = [det(A)]2 and det(A) = 1. 41. det(A) = det(AT ) = det(A) = (1)n (det A) =  det(A), so that det(A) = 0. We have used Facts 6.2.7 and 6.2.1a. 42. det(AT A) = det((QR)T QR) = det(RT QT QR) = det(RT Im R) = det(RT R) = det(RT ) det(R) Definition of A
m 2
Since columns of Q are orthonormal rii
i=1
Fact 6.2.4
= [det(R)] = Fact 6.2.7 Since R
2
>0
is triangular. vT wT [v w] = det vv vw vw ww = det v 2 vw vw w 2
43. det(AT A) = det = v
2
w
2
 (v w)2 0 by the CauchySchwarz inequality (Fact 5.1.11).
305
Chapter 6
ISM: Linear Algebra
44. a. We claim that v2 v3 vn = 0 if and only if the vectors v2 , . . . , vn are linearly independent. If the vectors v2 , . . . , vn are linearly independent, then we can find a basis x, v2 , . . . , vn of Rn (any vector x that is not in span (v2 , . . . , vn ) will do). Then x (v2 vn ) = det[x v2 vn ] = 0, so that v2 vn = 0. Conversely, suppose that v2 v3 vn = 0; say the ith component of this vector is nonzero. Then 0 = ei (v2 vn ) = det[ei v2 vn ], so that the vectors v2 , . . . , vn are linearly independent (being columns of an invertible matrix). 1 1 1 1 if i = 1 b. ith component of e2 e3 en = det ei e2 en = 0 if i > 1 1 1 1 so e2 e3 en = e1 . 1 c. vi (v2 v3 vn ) = det vi 1 d. Compare the 1 1 det ei v2 1 1 1 v2 1 1 v3 1 1 vn = 0 1
for any 2 i n since the above matrix has two identical columns. ith components of the two vectors: 1 1 1 1 1 v3 vn and det ei v3 v2 1 1 1 1 1 1 vn 1
The two determinants differ by a factor of 1 by Fact 6.2.1b, so that v2 v3 vn = v3 v2 vn .
2
e. det[v2 v3 vn v2 v3 vn ] = (v2 v3 vn )(v2 v3 vn ) = v2 vn
f. In Definition 6.1.1 we saw that the "old" cross product satisfies the defining equation of the "new" cross product: x (v2 v3 ) = det [ x v2 v3 ].
45. f (x) is a linear function, so f (x) is the coefficient of x (the slope). Expanding down the 1 2 3 4 0 2 3 4 first column, we see that the coefficient of x is  det = 24, so f (x) = 0 0 3 4 0 0 0 4 24.
306
ISM: Linear Algebra a 3 d a 1 d 46. a. det b 3 e = 3 det b 1 e = 21 c 3 f c 1 f
Section 6.2
Facts 6.2.1 a 3 d a 2(1) + 1 d a 2(1) d a 1 d b. det b 5 e = det b 2(2) + 1 e = det b 2(2) e + det b 1 e c 7 f c 2(3) + 1 f c 2(3) f c 1 f Fact 6.2.8 a 1 d a 1 d = 2 det b 2 e + det b 1 e = 2 11 + 7 = 29 c 3 f c 1 f Fact 6.2.1a
47. Yes! For example, T the first column.
x y
b = dx + by is given by the matrix [d b], so that T is linear in d
48. Since v2 , . . . , vn are linearly independent, T (x) = 0 only if x is a linear combination of the vi `s, (otherwise the matrix [x v2 vn ] is invertible, and T (x) = 0). Hence, the kernel of T is the span of v2 , . . . , vn , an (n  1)dimensional subspace of Rn . The image of T is the real line R (since it must be 1dimensional). 49. For example, 1 1 B = 0 1 0 0 we start with an upper triangular matrix B with det(B) = 13, such as can 1 1 . Adding the first row of B to both the second and the third to make 13 1 1 1 all entries nonzero, we end up with A = 1 2 2 . Note that det(A) = det(B) = 13. 1 1 14 Subtracting the second to last row from the last, we can make the last row into [ 0 0 0 1 ]. 307
50. There are many ways to do this problem; here is one possible approach:
Chapter 6
ISM: Linear Algebra
Now expanding along the last row we see that det(Mn ) = det(Mn1 ). Since det(M1 ) = 1 we can conclude that det(Mn ) = 1 for all n. 51. Notice that it takes n row swaps (swap row i with n + i for each i between 1 and n) to turn A into I2n . So, det(A) = (1)n det(I2n ) = (1)n .
52. a. We build B columnbycolumn: T 1 0 T 0 1 = d c b a = d c b . a
b. det(A) = ad  bc = det(B). The two determinants are equal. c. BA = AB = d c a c b d b a a c b da  bc = d 0 0 = (ad  bc)I2 . cb + ad 0 = (ad  bc)I2 also. cb + da
d b ad  bc = c a 0
d. Any vector u in the image of A will be of the form c1
b a . We note that + c2 d c a d b a ad  bc b B = = = 0. The same is true of B . Thus, c c a c ca + ac d anything in the image of A will be in the kernel of B. Since both matrices have a rank of 1, the dimensions of the kernel and image of each will be exactly 1. So, it must be that im(A) = ker(B).
d b + c2 . However, c a d ab + ba b a b . Thus, = 0. The same is true for A = we see that c bc + ad a c d by the same reasoning as above, the image of B will equal the kernel of A. Also, any vector u in the image of B will be of the form c1 e. A1 =
1 adbc
d b = c a
1 adbc B.
53. a. See Exercise 37. b. If A = a c b , then A1 = d
1 det(A)
d c
b a
has integer entries.
308
ISM: Linear Algebra a1 + tb1 a3 + tb3 a2 + tb2 a4 + tb4
2
Section 6.2 a1 a3 a2 , a4
54. f (t) = (det(A + tB))2  1 = B= b1 b3 b2 b4
det
 . 1 assuming A =
Then the determinant above is a polynomial of degree 2 so its square is a polynomial of degree 4. Hence f (t) is a polynomial of degree 4. Since A, A + B, A + 2B, A + 3B, A + 4B are invertible and their inverses have integer entries, by Exercise 53a, it follows that their determinants are 1. Hence f (0) = f (1) = f (2) = f (3) = f (4) = 0. Since f is a polynomial of degree 4 with at least 5 roots, it follows that f (t) = 0 for all t, in particular for t = 5, so det(A + 5B) = 1. Hence A + 5B is an invertible 2 2 matrix whose inverse has integer entries by Exercise 53b. 55. We start out with a preliminary remark: If a square matrix A has two equal rows, then D(A) = 0. Indeed, if we swap the two equal rows and call the resulting matrix B, then B = A, so that D(A) = D(B) = D(A), by property b, and D(A) = 0 as claimed. Next we need to understand how the elementary row operations affect D. Properties a and b tell us about how row multiplications and row swaps, but we still need to think about row additions. We will show that if B is obtained from A by adding k times the ith row to the j th , then D(B) = D(A). Let's label the row vectors of A by v1 , . . . vn . By linearity of D in the j th row (property c) we have . . . . . . vi vi . . . . D(B) = D = D(A) + kD = D(A). . . vj + kvi vi . . . . . .
Note that in the last step we have used the preliminary remark. Now, using the terminology introduced on Page 262, we can write D(A) = (1)s k1 k2 kr D(rref A).
Next we observe that D(rref A) = det(rref A) for all square matrices A. Indeed, if A is invertible, then rref(A) = In , and D(In ) = 1 = det(In ) by property c of function D. If A fails to be invertible, then D(rref A) = 0 = det(rref A) by linearity in the last row. It follows that D(A) = (1)s k1 k2 kr D(rref A) = (1)s k1 k2 kr det(rref A) = det(A) for all square matrices, as claimed. 309
Chapter 6
ISM: Linear Algebra
56. a. We show first that D is linear in the ith row. v1 v1 M . . . . . . x xM . . 1 D . = det M det . . . vn vn M A AM The entries in the ith row of AM are linear combinations of the components xi of the vector x, while the other entries of AM are constants. Therefore, det(AM ) is a linear 1 combination of the xi (expand along the ith row). Since det M is a constant, we have v1 . . . D x = c1 x1 + c2 x2 + + cn xn for some constants ci , as claimed. . . . vn v1 v1 . . . . . . v v i j . A = .  B = . . . . v j vi . . . . . . vn vn v1 M . . . vj M det . . . vi M . . . = 1 det M
b. Secondly, we need to show that D(B) = D(A) if B is obtained from A by a row swap:
D(B) =
1 det M
vn M
v1 M . . . vi M . det . . v M j . . .
vn M
= D(A).
c. The property D(In ) = 1 is obvious. 310
ISM: Linear Algebra It now follows from Exercise 41 that det(A) = D(A) = det(AM ) = det(A) det(M ).
det(AM ) det(M )
Section 6.2 and therefore
57. Note that matrix A1 is invertible, since det(A1 ) = 0. Now T y y = A1 y + A2 x = 0 when A1 y = A2 x, or, = [A1 A2 ] x x
y = A1 A2 x. This shows that for every x there is a unique y (that is, y is a function of 1 x); furthermore, this function is linear, with matrix M = A1 A2 . 1 58. Using the approach of Exercise 57, we have A1 = and M = A1 A2 = 1 1 8 . The function is 1 3 1 2 1 2 , , A2 = 4 3 3 7 y1 y2 = 1 8 1 3 x1 . x2
Alternatively, we can solve the linear system y1 + 2y2 + x1 + 2x2 3y1 + 7y2 + 4x1 + 3x2 Gaussian Elimination gives y1  x1 + 8x2 = 0 y2 + x1  3x2 = 0 59. det and y1 = x1  8x2 y2 = x1 + 3x2 =0 =0
0 0 1 0 = 0 = det , but these matrices fail to be similar. 0 0 0 0
60. We argue using induction on n. The base case (n = 2) is discussed on page 261. Now we assume that B is obtained from the n n matrix A by adding k times the pth row to the q th row. We will evaluate the determinant of B by expanding across the ith row (where i is neither p nor q). det(B) = n (1)i+j bij det(Bij ) j=1 = n (1)i+j aij det(Bij ) = n (1)i+j aij det(Aij ) = det(A) j=1 j=1 311
Chapter 6
ISM: Linear Algebra
Note that the (n  1) (n  1) matrix Bij is obtained from Aij by adding k times some row to another row. Now, det(Bij ) = det(Aij ) by induction hypothesis. 61. We follow the hint: = In C 0 A A C In C B D B D 0 A = A C A CA + AC B D B CB + AD
A B . So, det 0 AD  CB A C
= det(A) det(AD  CB).
Thus, det(In ) det(A) det det A C B D
= det(A) det(AD  CB), which leads to
= det(AD  CB), since det(A) = 0. In CA1 0 In A C B D A 0
62. a. We compute
B . Since the matrix CA1 B + D In 0 A B is invertible (its determinant is 1),the product CA1 In 0 CA1 B + D A B , namely, n. With A being invertible, this implies will have the same rank as C D that CA1 B + D = 0, or CA1 B = D, as claimed. =
b. Take determinants on both sides of the equation D = CA1 B from part (a) to find that det(D) = det(C)(det A)1 det(B), or det(A) det(D)  det(B) det(C) = 0, proving the claim.
6.3
1. By Fact 6.3.3, the area equals  det 2. By Fact 6.3.3 Area =
1 2
3 8  =   50 = 50. 7 2 = 1 50 = 25 2
det
3 8 7 2
1 3. Area of triangle = 2  det
6 1  = 13 (See Figure 6.1.) 2 4
1 2
4. Note that area of triangle =
b1  a 1 c1  a 1 . (See Figure 6.2.) b2  a 2 c2  a 2 a 1 b1 c1 a 1 b1  a 1 c1  a 1 On the other hand, det a2 b2 c2 = det a2 b2  a2 c2  a2 , by subtracting 1 1 1 1 0 0 the first column from the second and third. det 312
ISM: Linear Algebra
Section 6.3
Figure 6.1: for Problem 6.3.3.
Figure 6.2: for Problem 6.3.4. b1  a 1 b2  a 2
1 2
This, in turn, equals det
c1  a 1 , by expanding across the bottom row. c2  a 2 b1 b2 1 c1 c2 . 1
Therefore, area of triangle =
a1 det a2 1
1 1 5. The volume of the tetrahedron T0 defined by e1 , e2 , e3 is 3 (base)(height) = 6 .
Here we are using the formula for the volume of a pyramid. (See Figure 6.3.)
Figure 6.3: for Problem 6.3.5.
313
Chapter 6
ISM: Linear Algebra
The tetrahedron T defined by v1 , v2 , v3 can be obtained by applying the linear transformation with matrix [v1 v2 v3 ] to T0 .
1 Now we have vol(T ) =  det[v1 v2 v3 ]vol(T0 ) = 1  det[v1 v2 v3 ] = 6 V (v1 , v2 , v3 ). 6
Fact 6.3.8 and Page 282
Fact 6.3.5 a 1 b1 c1 6. From Exercise 5 we know that volume of tetrahedron = 1 det a2 b2 c2 , and Exer6 1 1 1 a 1 b1 c1 1 cise 4 tells us that area of triangle = 2 det a2 b2 c2 , so that area of tetrahedron = 1 1 1 1 (area of triangle). 3 We can see this result more directly if we think of the tetrahedron as an inverted pyramid whose base is the triangle and whose height is 1. (See Figure 6.4.)
Figure 6.4: for Problem 6.3.6. a1 b1 c1 The three vertices of the shaded triangle are a2 , b2 , c2 . 1 1 1 7. Area =
1 2
det
10 2 11 13
+
1 2
det
8 10 2 11
= 110. (See Figure 6.5.)
8. We need to show that both sides of the equation in Fact 6.3.4 give zero.  det(A) = 0 since A is not invertible. On the other hand, since A is not invertible, the vi will be linearly dependent, i.e., one of the vi will be redundant. This implies that vi = vi and vi = 0, so that the righthand side of the equation is 0, as claimed. 9. By Fact 6.3.3,  det[v1 v2 ] = area of the parallelogram defined by v1 and v2 . But v1 is the base of that parallelogram and v2 sin is its height, hence  det[v1 v2 ] = v1 v2 sin . 314
ISM: Linear Algebra
Section 6.3
Figure 6.5: for Problem 6.3.7.
10.  det(A) v1 v2 vn since  det(A) = v1 v2 vn and vi vi . The equality holds if vi = vi for all i, that is, if the vi `s are mutually perpendicular.
11. The matrix of the transformation T with respect to the basis v1 , v2 is B = that det(A) = det(B) = 12, by Fact 6.2.5.
3 0 , so 0 4
12. Denote the columns by v1 , v2 , v3 , v4 . From Fact 6.3.4 and Exercise 8 we know that  det(A) v1 v2 v3 v4 ; equality holds if the columns are orthogonal. Since the entries of the vi are 0, 1, and 1, we have vi 1 + 1 + 1 + 1 = 2. Therefore,  det A 16. To build an example where det(A) = 16 we want all 1's and 1's as entries, and the 1 1 1 1 1 1 1 1 columns need to be orthogonal. A little experimentation produces A = 1 1 1 1 1 1 1 1 (there are other solutions). Note that we need to check that det(A) = 16 (and not 16). 13. By Fact 6.3.7, the desired 2volume is 315
Chapter 6 1 1 1 1 1 1 det 1 1 2 3 4 1 1 2 = 3 4 1 1 1 1
ISM: Linear Algebra
det
4 10 = 20. 10 30
14. By Fact 6.3.7, the desired 3volume is 1 1 0 0 0 0 det 1 1 1 1 0 1 2 3 4 0 1 2 = 3 4 1 1 1 det 1 4 10 = 6. 1 10 30
15. If v1 , v2 , . . . , vm are linearly dependent and if A = [v1 vm ], then det(AT A) = 0 since AT A and A have equal and nonzero kernels (by Fact 5.4.2), hence AT A fails to be invertible. On the other hand, since the vi are linearly dependent, at least one of them will be redun dant. For such a redundant vi , we will have vi = vi and vi = 0, so that V (v1 , . . . , vm ) = 0, by Definition 6.3.6. This discussion shows that V (v1 , . . . , vm ) = 0 = det(AT A) if the vectors v1 , . . . , vm are linearly dependent. 16. False If T is given by A = 2I3 then  det(A) = 8. But if is the square defined by e1 , e2 in R3 (of area 1), then T () is the square defined by 2e1 , 2e2 and the area of T () is 4.
17. a. Let w = v1 v2 v3 . Note that w is orthogonal to v1 , v2 and v3 , by Exercise 6.2.44c. Then V (v1 , v2 , v3 , w) = V (v1 , v2 , v3 ) w = V (v1 , v2 , v3 ) w . Definition 6.3.6 b. By Exercise 6.2.44e, V (v1 , v2 , v3 , v1 v2 v3 ) =  det [ v1 v2 v3 v1 v2 v3 ]  =  det [ v1 v2 v3 v1 v2 v3 ]  = v1 v2 v3
2
.
c. By parts a and b, V (v1 , v2 , v3 ) = v1 v2 v3 . If the vectors v1 , v2 , v3 are linearly dependent, then both sides of the equation are 0, by Exercise 15 and Exercise 6.2.44a. 18. a. (See Figure 6.6.) 0 . q p 0 0 q p p cos(t) cos(t) and , the ellipse with semiaxes = 0 q sin(t) sin(t)
316
ISM: Linear Algebra
Section 6.3
Figure 6.6: for Problem 6.3.18a. (area of the ellipse) =  det(A)(area of the unit circle) = pq b. By Fact 6.3.8,  det(A) = area of the ellipse = area of the unit circle
ab
= ab so  det(A) = ab. 1 1 1 + sin(t) 2 ; 1 1
1 c. The unit circle consists of all vectors of the form x = cos(t) 2
its image is the ellipse consisting of all vectors 1 1 T (x) = cos(t) 2 2 + sin(t) 2 1 1 semimajor axis semiminor axis . (See Figure 6.7.)
19. det[v1 v2 v3 ] = v1 (v2 v3 ) = v1 v2 v3 cos where is the angle between v1 and v2 v3 so det[v1 v2 v3 ] > 0 if and only if cos > 0, i.e., if and only if is acute (0 ). 2 (See Figure 6.8.) 20. By Exercise 19, v1 , v2 , v3 constitute a positively oriented basis if and only if det[v1 v2 v3 ] > 0. Assume that v1 , v2 , v3 is such a basis. We want to show that Av1 , Av2 , Av3 is positively oriented if and only if det(A) > 0. We have det[Av1 Av2 Av3 ] = det(A[v1 v2 v3 ]) = det(A) det[v1 v2 v3 ] so since det [v1 v2 v3 ] > 0 by assumption, det [Av1 Av2 Av3 ] > 0 if and only if det(A) > 0. Hence A is orientation preserving if and only if det(A) > 0.
21. a. Reverses 317
Chapter 6
ISM: Linear Algebra
Figure 6.7: for Problem 6.3.18c.
Figure 6.8: for Problem 6.3.19. Consider v2 and v3 in the plane (not parallel), and let v1 = v2 v3 ; then v1 , v2 , v3 is a positively oriented basis, but T (v1 ) = v1 , T (v2 ) = v2 , T (v3 ) = v3 is negatively oriented. b. Preserves Consider v2 and v3 orthogonal to the line (not parallel), and let v1 = v2 v3 ; then v1 , v2 , v3 is a positively oriented basis, and T (v1 ) = v1 , T (v2 ) = v2 , T (v3 ) = v3 is positively oriented as well. c. Reverses 318
ISM: Linear Algebra
Section 6.3
The standard basis e1 , e2 , e3 is positively oriented, but T (e1 ) = e1 , T (e2 ) = e2 , T (e3 ) = e3 is negatively oriented. 3 7 1 , det(A) = 5, b = , so by Fact 6.3.9 4 11 3
# det "
22. Here A =
"
det
x=
1 7 3 11
5
= 2, y =
3 1 4 3
5
#
= 1.
23. Here A =
5 6
3 1 , det(A) = 17, b = , so by Fact 6.3.9 7 0 det 1 3 0 7 17
5 1 det 6 0 7 6 x1 = = , x2 = = . 17 17 17 2 3 0 8 24. Here A = 0 4 5 , det(A) = 146, b = 3 , so by Fact 6.3.9, 6 0 7 1
6 det6 4 2
x=
8 3 0 7 3 4 57 5 1 0 7
3 146
= 1, y =
6 det6 0 4
2
2
8 0 7 3 57 5 6 1 7
3 146
= 2, z =
6 det6 0 4
2
2 3 8 7 4 37 5 6 0 1
3 146
= 1.
25. By Fact 6.3.10, the ij th entry of adj(A) is given by (1)i+j det(Aji ), so since 1 0 1 1 0 A = 0 1 0 for i = 1, j = 1, we get (1)2 det = 1, and for i = 1, j = 2 we 0 1 2 0 1 0 1 get (1)3 det = 0, and so forth. 0 1 1 0 1 Completing this process gives adj(A) = 0 1 0 , hence by Fact 6.3.10, 2 0 1 1 0 1 1 0 1 1 1 A1 = det(A) adj(A) = 1 0 1 0 = 0 1 0 . 2 0 1 2 0 1
1 26. By Fact 6.3.10, A1 = det(A) adj(A), so if det(A) = 1, A1 = adj(A). If A has integer entries then (1)i+j det(Aji ) will be an integer for all 1 i, j n, hence adj(A) will have integer entries. Therefore, A1 will also have integer entries.
319
Chapter 6
ISM: Linear Algebra a b 1 , det(A) = a2 + b2 , b = , we get b a 0 =
a a2 +b2 , y
27. By Fact 6.3.9, using A = x = det 1 b 0 a
1 a2 +b2
= det
a 1 b 0
1 a2 +b2
=
b a2 +b2 ,
so x is positive, y is negative (since a, b > 0), and x decreases as b increases. 28. Here A =
"
I + G s a , det(A) = sh  ma, b = Ms + M m h a h
# #
so, by Fact 6.3.9
det
Y =
det
I + G Ms  M
shma
=
h(I +G)a(Ms M ) shma
=
h(I +G)+a(Ms M ) , sh+ma
r=
"
s I + G m Ms  M
shma
=
s(Ms M )m(I +G) shma
=
m(I +G)s(Ms M ) . sh+ma
29. By Fact 6.3.9,
6 det6 4 2 2
dx1 =
0 0 R2 de2 R1 R2
R1 1 R2
D
(1  ) 7 (1  )2 7 5 2  (1)
3
=
R1 R2 (1)2 de2 R2 (1)2 de2 D
dy1 =
6 det6 4 2
0 0 R2 de2
D
(1  ) 7 (1  )2 7 5 (1)2 
3 3
=
R2 de2 (R1 (1)2 +(1)) D
>0
dp =
6 det6 4
R1 R2
R1 1 R2
D
0 7 7 0 5 R2 de2
=
R1 R2 de2 D
> 0. 0 0 6 0. 5 3 0 1 5 2 . 5 1
18 30. Using the procedure outlined in Exercise 25, we find adj(A) = 12 2 6 31. Using the procedure outlined in Exercise 25, we find adj(A) = 3 4
0 0 0 1 0 0 1 0 32. Using the procedure outlined in Exercise 25, we find that adj(A) = . 0 0 1 0 1 0 0 0 320
ISM: Linear Algebra
Section 6.3
24 0 0 0 0 12 0 0 33. Using the procedure outlined in Exercise 25, we find that adj(A) = . 0 0 8 0 0 0 0 6 Note that the matrix adj(A) is diagonal, and the ith diagonal entry of adj(A) is the product of all ajj where j = i. 34. For an invertible n n matrix A, Aadj(A) = A(det(A)A1 ) = det(A)AA1 = det(A)In . The same is true for adj(A)A. 35. det(adj(A)) = det(det(A)A1 ). Taking the product det(A)A1 amounts to multiplying 1 each row of A1 by det(A), so that det(adj(A)) = (det A)n det(A1 ) = (det A)n det(A) = n1 (det A) . 36. adj(adjA) = adj(det(A)A1 ) = det(det(A)A1 )(det(A)A1 )1 = (det A)n det(A1 )(det(A)A1 )1
1 = (det A)n1 (det(A)A1 )1 = (det A)n1 det(A) (A1 )1
= (det A)n2 A. 37. adj(A1 ) = det(A1 )(A1 )1 = (det A)1 (A1 )1 = (adjA)1 . 38. adj(AB) = det(AB)(AB)1 = det(A)(det(B)B 1 )A1 = det(B)B 1 (det(A)A1 ) = adj(B)adj(A). 39. Yes, let S be an invertible matrix such that AS = SB, or SB 1 = A1 S. Multiplying both sides by det(A) = det(B), we find that S(det(B)B 1 ) = (det(A)A1 )S, or, S(adjB) = (adjA)S, as claimed. 40. The ij th entry of the matrix B of T is (ith component of T (ej )) = det(A(ej,i )), which is the ij th entry of adj(A) (see the first paragraph on Page 286 and Fact 6.3.10). Thus B = adj(A). 41. If A has a nonzero minor det(Aij ), then the n  1 columns of the invertible matrix Aij will be independent, so that the n  1 columns of A, minus the j th , will be independent as well. Thus, the rank of A (the dimension of the image) is at least n  1. Conversely, if rank(A) n  1, then we can find n  1 independent columns of A. The n (n  1) matrix consisting of those n  1 columns will have rank n  1, so that there will be exactly one redundant row (compare with Exercises 3.3.49 through 51). Omitting this redundant row produces an invertible (n  1) (n  1) submatrix of A, giving us a nonzero minor of A. 321
Chapter 6
ISM: Linear Algebra
42. By Definition 6.3.10, adj(A) = 0 if (and only if) all the minors Aji of A are zero. By Exercise 41, this is the case if (and only if) rank(A) n  2. 43. A direct computation shows that A(adjA) = (adjA)A = (det A)(In ) for all square matrices. Thus we have A(adjA) = (adjA)A = 0 for noninvertible matrices, as claimed. Let's write B = adj(A), and let's verify the equation AB = (det A)(In ) for the diagonal entries; the verification for the offdiagonal entries is analogous. The ith diagonal entry of AB is n ith [ith row of A] column = ai1 b1i + + ain bni = aij bji . j=1 of B Since B is the adjunct of A, bji = (1)j+i det(Aij ).
n
So, our summation equals
aij (1)i+j det(Aij )
j=1
which is our formula for Laplace expansion across the ith row, and equals det(A), proving our claim for the diagonal entries. 44. The equation A(adjA) = 0 from Exercise 43 means that im(adjA) is a subspace of ker(A). Thus rank(adjA) = dim(im(adjA)) dim(kerA) = n  rank(A) = n  (n  1) = 1, implying that rank(adjA) 1. Since adj(A) = 0, by Exercise 42, we can conclude that rank(adjA) = 1. 45. Let A = b . So, a = d and a a b b = c. Thus, the equation AT = adj(A) holds for all matrices of the form . b a a c b . We want AT = adj(A), or d a b c d = f (x, y)dA =
2 2
d c
46. In the simple case when f (x, y) = 1 we have and g(u, v)dA =
1 1
dA = area of 2 =  det M 
dA = area of 1 = 1, so that
2
f (x, y)dA =  det M 
g(u, v)dA.
1
This formula holds, in fact, for any continuous function f (x, y); see an introductory text in multivariable calculus for a justification. 47. Note that x1 x2 is the area of the triangle OP1 P2 , where O denotes the origin. y1 y2 This is likewise true for onehalf the second matrix. However, because of the reversal
1 2
det
322
ISM: Linear Algebra
1 2
Section 6.3
x3 x4 is negative the area of the triangle OP3 P4 ; likewise for y3 y4 the last matrix. (See the discussion on Page 277.) Finally, note that the area of the quadrilateral P1 P2 P3 P4 is equal to: in orientation, det the area of triangle OP1 P2 + area of triangle OP2 P3  area of triangle OP3 P4  area of triangle OP4 P1 . 48. In what follows, we will freely use the fact that an invertible linear transformation L from R2 to R2 maps an ellipse into an ellipse (see Exercise 2.2.52). Now consider a linear transformation L that transforms our 345 right triangle R into an equilateral triangle T . If we place the vertices of the right triangle R at the points 0 0 , 4 0 , , and the vertices of the equilateral triangle T at 0 3
1 2
0 2 1 , , , 3 0 0
1 . 2 3
then the transformation L has the matrix A =
0
1 3 1 3
, with det(A) =
According to the hint, L will map the largest ellipse E inscribed into R into the circle C inscribed into T . The Figure 6.9 illustrates that the radius of C is tan(/6) = 1/ 3, so that the area of C is /3. Using the interpretation of the determinant as an expansion 2 fact, we find that (area of C) = (det A)(area of E), or (area of E) = area of C = 3 3.6 det(A)
1 2 A= R 2 0 L L1 1
1
[0 [ 3
1 3 1 3
[ 13 [
2 3 3 C T
2 0
A =
1
2
3
[4[ 0
6
r= 1 3
[2[ 0
Figure 6.9: for Problem 6.3.48 and Problem 49. 49. We will use the terminology introduced in the solution of Exercise 48 throughout. Note 2 2/ 3 that the transformation L1 , with matrix A1 = , maps the circle C (with 0 3 323
Chapter 6
ISM: Linear Algebra
1 3
cos of C, sin 2 4 and find the maximal value M and the minimal value m of A1 v = 3 + 1 sin2  9 8 25 1 4 (sin )(cos ) = 18  18 (cos 2)  3 3 (sin 2) (we are taking the square to facilitate 3 3 the computations). Then M and m will be the lengths of the semiaxes of E. The function above is sinusoidal with average value Thus M = M= m=
25+ 193 18
radius 1/ 3) into the ellipse E. Now consider a radial vector v =
and m =
25 193 , 18
25 18
and amplitude
1 182
+
16 27
=
193 18 .
so that the length of the semimajor axis of E is
25+ 193 18 25 193 18
1.47, and for the semiminor axis we get 0.79.
True or False
1. T, by Fact 6.1.6 (a diagonal matrix is triangular as well) 2. T, by Fact 6.2.1b. 3. T, by Definition 6.1.1 4. F; We have det(4A) = 44 det(A), by Fact 6.2.1a. 5. F; Let A = B = I5 , for example 6. T; We have det(A) = (1)6 det(A) = det(A), by Fact 6.2.1a. 7. F; In fact, det(A) = 0, since A fails to be invertible 8. F; The matrix A fails to be invertible if det(A) = 0, by Fact 6.2.2. 9. T, by Fact 6.2.1a, applied to the columns. 10. T, by Fact 6.2.4 11. T, by Fact 6.2.5. 12. F, by Fact 6.3.1. The determinant can be 1. 13. T, by Fact 6.2.4. 14. F; The second and the fourth column are linearly dependent. 324
ISM: Linear Algebra
True or False
15. T; The determinant is 0 for k = 1 or k = 2, so that the matrix is invertible for all positive k. 16. F; Expand down the first column to see that det(A) =  det(I3 ) = 1. 17. T; It would be tedious to compute the exact value of the determinant, but we can show that the determinant is nonzero without finding its exact value. One method is to show that the determinant is an odd integer. Expanding down the third 5 4 8 column, we see that det(A) = 3 det 100 9 7 + even terms 6 5 100 = 3(7 det 5 4 + even terms ) + even terms 6 5
= 21 + even terms = odd, as claimed. Alternatively, we can use the permutation formula for the determinant (Fact 6.2.10), and argue that one of the 24 terms is very large (namely 1004 = 108 ) compared to the other terms [those are less than (1002 )(102 ) = 106 each in absolute value], so that the determinant turns out to be positive. (In fact, this determinant is 97,763,383.) 18. F; The correct formula is det(A1 ) = 19. T; Matrix A is invertible. 20. T; Any nonzero noninvertible matrix A will do.
21. T, by Fact 6.3.4, since vi vi = 1 for all column vectors vi . 1 det(AT ) ,
by Facts 6.2.6 and 6.2.7.
22. T; We have det(A) = det(rref A) = 0. 23. F; Let A = 3 2 , for example. 5 3
24. F; Let A = 2I2 , for example 1 1 1 1 1 1 1 1 25. T; Let A = . The column vectors of A are orthogonal and they all 1 1 1 1 1 1 1 1 have length 2. 26. F; Let A = 8 0 0
1 2
and v =
1 , for example. 0 325
Chapter 6
ISM: Linear Algebra
27. F; In fact, det(A) = det[u v w] =  det[v u w] = v (u w). We have used Fact 6.2.1b and Definition 6.1.1. 1 0 and 0 1 1 1 29. F; Note that det 0 1 1 0 28. T; Let A = B= 1 0 , for example. 0 1
30. T, by Fact 6.3.10
0 1 = 2. 1
33. F; Let A = I2 and B = I2 , for example.
31. F; Let A = 2I2 , for example. 0 1 0 0 0 0 0 1 32. F; Let A = , for example. 0 0 0 1 0 0 1 0
34. T; Note that det(B) =  det(A) < det(A), so that det(A) > 0. 35. T; Let's do Laplace expansion along the first row, for example (see Fact 6.1.5).
n
Then det(A) =
j=1
(1)1+j a1j det(A1j ) = 0. Thus det(A1j ) = 0 for at least one j, so that
A1j is invertible. 36. T; Note that det(A) and det(A1 ) are both integers, and (det A)(det A1 ) = 1. This leaves only the possibilities det(A) = det(A1 ) = 1 and det(A) = det(A1 ) = 1. 37. T, since adj(A) = (det A)(A1 ), by Fact 6.3.10. 38. F; Note that det(A2 ) = (det A)2 cannot be negative, but det(I3 ) = 1. 39. F; Note that det(S 1 AS) = det(A) but det(2A) = 23 (det A) = 8(det A). 40. F; Note that det(S T AS) = (det S)2 (det A) and det(A) = (det A) have opposite signs. 41. T; We can use induction on n to show that det(A) is an odd integer. If we expand det(A) down the first column, then the first term, a11 det(A11 ), will be odd, since a11 is odd, and det(A11 ) is odd by the induction hypothesis. However, all the other terms ai1 det(Ai1 ), where i > 1, will be even, since ai1 is even in this case. Thus, det(A) is odd as claimed, and A is invertible. 2 1 1 42. F; Let A = 1 2 1 , for example 1 5 2 326
ISM: Linear Algebra a c b d
True or False 0 0 0 0 ; if b = 0, let B = ; if c = 0, let 0 1 1 0
43. T; Let A = B=
. If a = 0, let B =
0 1 1 0 , and if d = 0, let B = . 0 0 0 0
44. T; Use Gaussian elimination for the first column only to transform A into a matrix of the form 1 1 1 1 0 B= 0 0 Note that det(B) = det(A) or det(B) = (det A). The stars in matrix B all represent numbers (1) (1), so that they are 2, 0, or 2. Thus the determinant of the 3 3 matrix M containing the stars is divisible by 8, since each of the 6 terms in Sarrus' rule is 8, 0 or 8. Now perform Laplace expansion down the first column of B to see that det(M ) = det(B) = +/  det(A). 45. T; A(adjA) = A(det(A)A1 ) = det(A)In = det(A)A1 A = adj(A)A. 1 2 4 46. T; Laplace expansion along the second row gives det(A) = k det 8 9 7 + C = 0 0 5 35k + C, for some constant C (we need not compute that C = 259). Thus A is invertible except for k = C (which turns out to be 259 = 37 = 7.4). 35 35 5 1 0 0 1 0 0 47. F; A = 0 0 1 and B = 0 1 0 are both orthogonal and det(A) = 0 1 0 0 0 1 det(B) = 1. However, AB = BA.
327
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