32 by exercise 31 we need to compute q ij a i a j

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32. By Exercise 31, we need to compute Q i>j ( a i - a j ) where a 0 = 1 , a 1 = 2 , a 2 = 3 , a 3 = 4 , a 4 = 5 so Q i>j ( a i - a j ) = (2 - 1)(3 - 1)(3 - 2)(4 - 1)(4 - 2)(4 - 3)(5 - 1)(5 - 2)(5 - 3)(5 - 4) = 288. 303
Chapter 6 ISM: Linear Algebra 33. Think of the i th column of the given matrix as a i 1 a i a 2 i . . . a n - 1 i , so by Fact 6.2.1a, the determi- nant can be written as ( a 1 a 2 · · · a n ) det 1 1 · · · 1 a 1 a 2 · · · a n a 2 1 a 2 2 · · · a 2 n . . . . . . . . . a n - 1 1 a n - 1 2 · · · a n - 1 n . The new determinant is a Vandermonde determinant (see Exercise 31), and we get n Y i =1 a i Y i>j ( a i - a j ) . 34. a. The hint pretty much gives it away. Since the columns of matrix B - I n are in the kernel of [ I n M ], we have [ I n M ] B - I n = B - M = 0, and M = B , as claimed. b. If B = A - 1 we get rref[ A . . . I n ] = [ I n . . . A - 1 ] which tells us how to compute A - 1 (see Fact 2.3.5). 35. x 1 x 2 must satisfy det 1 1 1 x 1 a 1 b 1 x 2 a 2 b 2 = 0 , i.e., must satisfy the linear equation ( a 1 b 2 - a 2 b 1 ) - x 1 ( b 2 - a 2 ) + x 2 ( b 1 - a 1 ) = 0 . We can see that x 1 x 2 = a 1 a 2 and x 1 x 2 = b 1 b 2 satisfy this equation, since the matrix has two identical columns in these cases. 36. Expanding down the first column we see that the equation has the form A - Bx 1 + Cx 2 - D ( x 2 1 + x 2 2 ) = 0. If D 6 = 0 this equation defines a circle; otherwise it is a line. From Exercise 35 we know that D = 0 if and only if the three given points a 1 a 2 , b 1 b 2 , c 1 c 2 are collinear. Note that the circle or line runs through the three given points. 304
ISM: Linear Algebra Section 6.2 37. Applying Fact 6.2.4 to the equation AA - 1 = I n we see that det( A ) det( A - 1 ) = 1 . The only way the product of the two integers det( A ) and det( A - 1 ) can be 1 is that they are both 1 or both -1. Therefore, det( A ) = 1 or det( A ) = - 1 . 38. det( A T A ) = det( A T ) det( A ) = [det( A )] 2 = 9 Fact 6.2.4 Fact 6.2.7 39. det( A T A ) = det( A T ) det( A ) = [det( A )] 2 > 0 Fact 6.2.4 Fact 6.2.7 40. By Exercise 38, det( A T A ) = [det( A )] 2 . Since A is orthogonal, A T A = I n so that 1 = det( I n ) = det ( A T A ) = [det( A )] 2 and det( A ) = ± 1. 41. det( A ) = det( A T ) = 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( A T A ) = det(( QR ) T QR ) = det( R T Q T QR ) = det( R T I m R ) = det( R T R ) = det( R T ) det( R ) Definition of A Since columns of Q are orthonormal Fact 6 . 2 . 4 = [det( R )] 2 = m Y i =1 r ii ! 2 > 0 Fact Since R 6.2.7 is triangular. 43. det( A T A ) = det ~v T ~w T [ ~v ~w ] = det ~v · ~v ~v · ~w ~v · ~w ~w · ~w = det k ~v k 2 ~v · ~w ~v · ~w k ~w k 2 = k ~v k 2 k ~w k 2 - ( ~v · ~w ) 2 0 by the Cauchy-Schwarz inequality (Fact 5.1.11). 305
Chapter 6 ISM: Linear Algebra 44. a. We claim that ~v 2 × ~v 3 × · · · × ~v n 6 = ~ 0 if and only if the vectors ~v 2 , . . . ,~v n are linearly independent. If the vectors ~v 2 , . . . ,~v n are linearly independent, then we can find a basis ~x,~v 2 , . . . ,~v n of R n (any vector ~x that is not in span ( ~v 2 , . . . ,~v n ) will do). Then ~x · ( ~v 2 × · · · × ~v n ) = det[ ~x~v 2 · · · ~v n ] 6 = 0, so that ~v 2 × · · · × ~v n 6 = ~ 0. Conversely, suppose that ~v 2 × ~v 3 × · · · × ~v n 6 = 0; say the i th component of this vector is nonzero. Then 0 6 = ~ e i · ( ~v 2 × · · · × ~v n ) = det[ ~ e i ~v 2 · · · ~v n ], so that the vectors ~v 2 , . . . ,~v n are linearly

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