# 24 we can write f x x 3 x x x 2 1 x x 1 x 1 the

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24. We can write f ( x ) = x 3 - x = x ( x 2 - 1) = x ( x - 1)( x + 1). The equation f ( x ) = 0 has three solutions, x = 0 , 1 , - 1, so that f ( x ) fails to be invertible. 25. Invertible, with inverse x 1 x 2 = 3 y 1 y 2 26. Invertible, with inverse x 1 x 2 = 3 y 2 - y 1 y 1 88
ISM: Linear Algebra Section 2.3 Figure 2.40: for Problem 2.3.23. 27. This transformation fails to be invertible, since the equation x 1 + x 2 x 1 x 2 = 0 1 has no solution. 28. We are asked to find the inverse of the matrix A = 22 13 8 3 - 16 - 3 - 2 - 2 8 9 7 2 5 4 3 1 . We find that A - 1 = 1 - 2 9 - 25 - 2 5 - 22 60 4 - 9 41 - 112 - 9 17 80 222 . T - 1 is the transformation from R 4 to R 4 with matrix A - 1 . 29. Use Fact 2.3.3: 1 1 1 1 2 k 1 4 k 2 - I - I 1 1 1 0 1 k - 1 0 3 k 2 - 1 - II - 3( II ) 1 0 2 - k 0 1 k - 1 0 0 k 2 - 3 k + 2 The matrix is invertible if (and only if) k 2 - 3 k + 2 = ( k - 2)( k - 1) = 0, in which case we can further reduce it to I 3 . Therefore, the matrix is invertible if k = 1 and k = 2. 30. Use Fact 2.3.3: 0 1 b - 1 0 c - b - c 0 ----→ I II - 1 0 c 0 1 b - b - c 0 ----→ ÷ ( - 1) 1 0 - c 0 1 b - b - c 0 + b ( I ) + c ( II ) -→ 1 0 - c 0 1 b 0 0 0 This matrix fails to be invertible, regarless of the values of b and c . 31. Use Fact 2.3.3; first assume that a = 0. 89
Chapter 2 ISM: Linear Algebra 0 a b - a 0 c - b - c 0 swap : I II - a 0 c 0 a b - b - c 0 ÷ ( - a ) 1 0 - c a 0 a b - b - c 0 + b ( I ) 1 0 - c a 0 a b 0 - c - bc a ÷ a 1 0 - c a 0 1 b a 0 - c - bc a + c ( II ) 1 0 - c a 0 1 b a 0 0 0 Now consider the case when a = 0: 0 0 b 0 0 c - b - c 0 swap : I III - b - c 0 0 0 c 0 0 b : The second entry on the diagonal of rref will be 0. It follows that the matrix 0 a b - a 0 c - b - c 0 fails to be invertible, regardless of the values of a, b , and c . 32. Use Fact 2.3.6. If A = a b c d is a matrix such that ad - bc = 1 and A - 1 = A , then A - 1 = 1 ad - bc d - b - c a = d - b - c a = a b c d , so that b = 0 , c = 0, and a = d . The condition ad - bc = a 2 = 1 now implies that a = d = 1 or a = d = - 1. This leaves only two matrices A , namely, I 2 and - I 2 . Check that these two matrices do indeed satisfy the given requirements. 33. Use Fact 2.3.6. The requirement A - 1 = A means that - 1 a 2 + b 2 - a - b - b a = a b b - a . This is the case if (and only if) a 2 + b 2 = 1. 34. a. By Fact 2.3.3, A is invertible if (and only if) a, b , and c are all nonzero. In this case, A - 1 = 1 a 0 0 0 1 b 0 0 0 1 c . 90
ISM: Linear Algebra Section 2.3 b. In general, a diagonal matrix is invertible if (and only if) all of its diagonal entries are nonzero. 35. a. A is invertible if (and only if) all its diagonal entries, a, d , and f , are nonzero. b. As in part (a): if all the diagonal entries are nonzero. c. Yes, A - 1 will be upper triangular as well; as you construct rref[ A . . . I n ], you will perform only the following row operations: divide rows by scalars subtract a multiple of the j th row from the i th row, where j > i .
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