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

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Math 2940 Solutions, Fall 2011 Section 2 . 7 1 ) A T = 1 9 0 3 . We have 1 0 | 1 0 9 3 | 0 1 1 0 | 1 0 0 3 | - 9 1 1 0 | 1 0 0 1 | - 3 1 / 3 so A - 1 = 1 0 - 3 1 / 3 and ( A - 1 ) T = 1 - 3 0 1 / 3 . A similar computation for ( A T ) - 1 gives ( A T ) - 1 = ( A - 1 ) T . This is not an accident! For A = 1 c c 0 , A T = 1 c c 0 = A . We have 1 c | 1 0 c 0 | 0 1 1 c | 1 0 0 - c 2 | - c 1 1 c | 1 0 0 1 | 1 /c - 1 /c 2 1 0 | 0 1 /c 0 1 | 1 /c - 1 /c 2 so A - 1 = 0 1 /c 1 /c - 1 /c 2 and ( A - 1 ) T = A . As before, ( A T ) - 1 = ( A - 1 ) T . Finally, all this depends on c = 0. If c = 0, then A - 1 does not exist. 7 ) (a) False. Suppose all the blocks are 2 × 2. Let A = 0 0 1 0 . b) False. Let A = 1 1 1 0 and B = 0 1 1 1 so AB = 1 2 0 1 is not symmetric. c) True. Say A = A T . Then inverting both sides of this inequality gives A - 1 = ( A T ) - 1 where the last term is ( A - 1 ) T . d) True. ( ABC ) T = ( A ( BC )) T = ( BC ) T A T = C T B T A T = CBA . 13 ) (a) 0 0 1 1 0 0 0 1 0 works. b) 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 1 4 = 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 1 = I . 17 ) (a) 0 0 0 1 .

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b) 0 1 1 0 . Indeed, suppose 0 1 1 0 = a 0 b c x y 0 z = ax ay cx + bx cy + bz . Then we must have ax = 0 and ay = 1 so x = 0 and 0 1 1 0 = 0 ay 0 cy + bz , an imposibility. No LU factorization exists.
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