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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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