28_hwc1SolnsODDA

# 28_hwc1SolnsODDA - and p ( A ) is its inverse. (b) The two...

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(b) Let A and B be the matrices from problem 3.30. Find the inverses of A and B using part (a) of this problem. SOLUTION (a) Write q ( x ) as q ( x ) = a 0 + a 1 x + a 2 x 2 + ··· + a k x k . Notice that q (0) = a 0 , so that q (0) - q ( x ) = - a 1 x - a 2 x 2 - ··· - a k x k , and hence p ( x ) = q (0) - q ( x ) xq (0) = - a 1 a 0 - a 2 a 0 x - ··· - a k a 0 x k - 1 , which is polynomial of degree k - 1. To handle the next part, we note that, by the deﬁnition of p ( x ), xp ( x ) = p ( x ) x = 1 - q ( x ) q (0) , which is a polynomial of degree k . Hence, Ap ( A ) = p ( A ) A = I - 1 q (0) q ( A ) = I , since q ( A ) = 0. Therefore, p ( A ) is both a left and a right inverse of A . Thus, A is invertible,
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Unformatted text preview: and p ( A ) is its inverse. (b) The two matrices from exercise 3.30 were A = 4 2-2 3 and B = 2-2 2 1 1 1 . We found there that with q ( x ) = x 2-7 x + 16, q ( A ) = 0. In this case, q (0) = 16, and q (0)-q ( x ) xq (0) = 1 16 (7-x ) . Therefore. by part (a) , A is invertible, and A-1 = 1 16 (7 I-A ) = 1 16 3-2 2 4 . 21 /september/ 2005; 22:09 29...
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## This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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