solutions-manual.21-30 - Exercises 21 7 Let A = p u2212 1...

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Exercises 21 7. Let A = p 1 1 3 3 P . Find all 2 × 2 matrices, B such that AB = 0 . p 1 1 3 3 Pp x y z w P = p x z w y 3 x + 3 z 3 w + 3 y P = p 0 0 0 0 P and so you have to solve Solution is: [ w = y,x = z ] So the matrices are of the form p x y x y P . 8. Let x = ( 1 , 1 , 1) and y = (0 , 1 , 2) . Find x T y and xy T if possible. 0 1 2 0 1 2 0 1 2 , 1 9. Let A = p 1 2 3 4 P ,B = p 1 2 3 k P . Is it possible to choose k such that AB = BA ? If so, what should k equal? p 1 2 3 4 1 2 3 k P = p 7 2 k + 2 15 4 k + 6 P p 1 2 3 k 1 2 3 4 P = p 7 10 3 k + 3 4 k + 6 P Thus you must have 3 k + 3 = 15 2 k + 2 = 10 , Solution is: [ k = 4] 10. Let A = p 1 2 3 4 P = p 1 2 1 k P . Is it possible to choose k such that AB = ? If so, what should k equal? p 1 2 3 4 1 2 1 k P = p 3 2 k + 2 7 4 k + 6 P p 1 2 1 k 1 2 3 4 P = p 7 10 3 k + 1 4 k + 2 P However, 7 n = 3 and so there is no possible choice of k which will make these matrices commute. 11. In 5.1 - 5.8 describe A and 0 . To get A, just replace every entry of A with its additive inverse. The 0 matrix is the one which has all zeros in it. 12. Let A be an n × n matrix. Show A equals the sum of a symmetric and a skew symmetric matrix. ( M is skew symmetric if M = M T . M is symmetric if M T = M .) Hint: Show that 1 2 ( A T + A ) is symmetric and then consider using this as one of the matrices. A = A + A T 2 + A - A T 2 . 13. Show every skew symmetric matrix has all zeros down the main diagonal. The main diagonal consists of every entry of the matrix which is of the form a ii . It runs from the upper left down to the lower right. If A = A T , then a ii = a ii and so each a ii = 0. 14. Suppose M is a 3 × 3 skew symmetric matrix. Show there exists a vector Ω such that for all u R 3 M u = Ω × u Saylor URL: The Saylor Foundation
22 Exercises Hint: Explain why, since M is skew symmetric it is of the form M = 0 ω 3 ω 2 ω 3 0 ω 1 ω 2 ω 1 0 where the ω i are numbers. Then consider ω 1 i + ω 2 j + ω 3 k . Since M is skew symmetric, it is of the form mentioned above. Now it just remains to verify that Ω ω 1 i + ω 2 j + ω 3 k is of the right form. But v v v v v v i j k ω 1 ω 2 ω 3 u 1 u 2 u 3 v v v v v v = 2 u 3 3 u 2 1 u 3 + 3 u 1 + 1 u 2 2 u 1 . In terms of matrices, this is ω 2 u 3 ω 3 u 2 ω 1 u 3 + ω 3 u 1 ω 1 u 2 ω 2 u 1 . If you multiply by the matrix, you get 0 ω 3 ω 2 ω 3 0 ω 1 ω 2 ω 1 0 u 1 u 2 u 3 = ω 2 u 3 ω 3 u 2 ω 3 u 1 ω 1 u 3 ω 1 u 2 ω 2 u 1 which is the same thing. 15. Using only the properties 5.1 - 5.8 show A is unique. Suppose B also works. Then A = A + ( A + B ) = ( A + A ) + B = 0 + B = B 16. Using only the properties 5.1 - 5.8 show 0 is unique. Suppose 0 is another one. Then 0 = 0 + 0 = 0 .

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