# 5.2 - Then we can row reduce the A part and D part of our...

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Math 2940 Solutions, Fall 2011 Section 5 . 2 12 ) C = 3 2 1 2 4 2 1 2 3 and AC T = 2 - 1 0 - 1 2 - 1 0 - 1 2 3 2 1 2 4 2 1 2 3 = 4 0 0 0 4 0 0 0 4 so C T = 4 A - 1 . 17 ) Expanding on the last row, det ( B 4 ) = - det 1 - 1 0 - 1 2 0 0 - 1 - 1 +2 det 1 - 1 0 - 1 2 - 1 0 - 1 2 = 2 | B 3 |-| B 2 | = 2 - 1 = 1 . 20 ) | G 2 | = - 1, G 3 = 2. 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 0 1 1 1 1 0 1 1 0 1 - 1 0 0 1 0 - 1 0 1 1 1 1 0 1 1 0 1 - 1 0 0 0 1 - 1 . Now we put the first row on the bottom. This is the same as doing 3 row switches so 1 0 1 1 0 1 - 1 0 0 0 1 - 1 0 1 1 1 1 0 1 1 0 1 - 1 0 0 0 1 - 1 0 0 2 1 1 0 1 1 0 1 - 1 0 0 0 1 - 1 0 0 0 3 so our determinant is ( - 1) 3 3 = - 3. We guess that G n = ( - 1) n - 1 n . 23 ) (a) We think about the BIG FORMULA. Whenever there is an entry from the B part of A B 0 D , the product involving this entry must also have entries from the third and fourth rows and one of these must be in the first or second columns. That entry is 0 so B does not enter into computing the determinant. We may as well assume B is the zero matrix. Then we can row reduce the A part and

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Unformatted text preview: Then we can row reduce the A part and D part of our block matrix independently so its determinant is det ( A ) det ( D ). (b) Consider 1 0 0 1 r 0 1 1 0 r 1 0 0 1 . Here B and C both have determinant 0 while A and D have determinant r , so | A || D | - | C || B | = r 2 . The ﬁrst and last rows of 1 0 0 1 r 0 1 1 0 r 1 0 0 1 are the same, so row reduction gives a row of zeros and the matrix has determinant 0. As long as r 6 = 0 the values are diﬀerent. (c) In the example above, AD-CB = ± 1 0 r ²± r 0 1 ²-± 1 0 1 0 ²± 0 1 0 1 ² = ± r r ²-± 0 1 0 1 ² = ± r-1 r-1 ² which has determinant r ( r-1) and is nonzero as long as r 6 = 0 , 1. But our big 4 × 4 determinant is 0....
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