2.7 - Math 2940 Solutions Fall 2011 Section 2.7 19 We have...

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Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 2.7 19 . We have 03 1 ) AT = 10|10 93|01 → 10| 10 0 3 | −9 1 → 10| 1 0 0 1 | −3 1/3 1 0 1 −3 and (A−1 )T = . A similar computation for (AT )−1 gives −3 1/3 0 1/3 (AT )−1 = (A−1 )T . This is not an accident! 1c 1c For A = , AT = = A. We have c0 c0 so A−1 = 1c|10 c0|01 → 1c| 1 0 0 1 | 1/c −1/c2 1 c| 10 2 0 −c | −c 1 → → 10| 0 1/c 0 1 | 1/c −1/c2 0 1/c and (A−1 )T = A. As before, (AT )−1 = (A−1 )T . Finally, all this 1/c −1/c2 depends on c = 0. If c = 0, then A−1 does not exist. so A−1 = 00 . 10 11 01 12 b) False. Let A = and B = so AB = is not symmetric. 10 11 01 c) True. Say A = AT . Then inverting both sides of this inequality givesA−1 = (AT )−1 where the last term is (A−1 )T . d) True. (ABC )T = (A(BC ))T = (BC )T AT = C T B T AT = CBA. 7) (a) False. Suppose all the blocks are 2 × 2. Let A = 13) (a) 0 1 b) 0 0 17) (a) 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 1 0 works. 0 4 00 1 0 = 0 1 00 00 . 01 1 0 0 0 0 0 = I. 0 1 b) 01 . Indeed, suppose 10 01 10 a0 bc = xy 0z = ax ay cx + bx cy + bz . Then we must have ax = 0 and ay = 1 so x = 0 and 01 10 = 0 ay 0 cy + bz , an imposibility. No LU factorization exists. c) So we cheat and set −1 0 0 −1 10 11 A= 11 01 = −1 −1 −1 −2 . a2 ab ab b2 + c2 . Suppose −1 −1 −1 −2 = a0 bc ab 0c = Then a2 = −1, which does not happen in the real numbers. We cannot write A = LLT . 0 1 28) Yes. Consider 2 3 1 2 3 0 2 3 0 1 3 0 . 1 2 39) (a,b) Given A, set X = .5(A + AT ) and Y = .5(A − AT ). Then A = X + Y and X T = .5(AT + A) = X and Y T = .5(AT − A) = −Y . ...
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2.7 - Math 2940 Solutions Fall 2011 Section 2.7 19 We have...

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