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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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