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Unformatted text preview: Solutions to Homework #9, MAT1341A Question # 1: (3+3=6points) Compute the determinants of the following matrices. (a) A = 3 1 2 3 5 2 1 4 Solution: We have several options. Method 1: By direct cofactor expansion, on row 1: det( A ) = 3 det 3 5 1 4 ( 1) det 2 5 2 4 = 3(7) + (18) = 39 Method 2: First do some row reduction, and keep track of steps: A 3 2 R 2 + R 1 ∼ R 2 + R 3  11 / 2 15 / 2 2 3 5 4 9 R 1 ↔ R 2 ∼ 2 3 5 11 / 2 15 / 2 4 9  2 / 11 R 2 ∼ 2 3 5 1 15 / 11 4 9  4 R 2 + R 3 ∼ 2 3 5 1 15 / 11 39 / 11 = R which is triangular and has determinant equal to the product of the diagonal entries, so det( R ) = 78 / 11. Now we have det( R ) = ( 1)( 2 / 11) det( A ) because of the row interchange and the scaling, and so det( A ) = 39 Method 3: Use a clever combination of techniques: det( A ) = det( A T ) = det 3 2 2 1 3 1 5 4 = det 11 1 1 3 1 5 4 = ( 1) det 11 1 5 4 = 44 5 = 39 (we did a row reduction step: 3R2+R1, which doesn’t change the determinant). (b) B =  3 1 5 1 1 3 2 3 4 1 2 3 4 Solution: We could do a direct cofactor expansion along any row or column; and certainly we would try to use a column with the most zeros; in this case, 1. Say we choose column 4; then recalling the signs on the cofactors are given by the formula ( 1) i + j we have det( A ) = det 1 3 3 4 2 3 4 + 2 det  2 1 5 3 4 2 3 4  det  3 1 5 1 3 3 4 and then you keep expanding. However, this involves a lot of calculation and is therefore quite errorprone; you would be much better off doing a few simple row reduction steps to create more zeros. For instance, you could do 2 R 1 + R 2 and R 1 + R 3; and that would zero out all but one entry in column 4, so instead of three 3 ×...
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 Winter '10
 DAIGLE
 Linear Algebra, Determinant, Matrices, #, Eigenvalue, eigenvector and eigenspace, Det

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