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hw9sol - Math 221 6.1 2 Compute the determinant of A = det...

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Math 221 Problem Set 8 Solutions 6.1: 2 Compute the determinant of A = 2 3 4 5 . Is A invertible? det 2 3 4 5 = 2 * 5 - 3 * 4 = 10 - 12 = - 2. 6.1: 22 Compute the determinant of A = cos k 1 - sin k 0 2 0 sin k 0 cos k . For which values of k is A invertible? det cos k 1 - sin k 0 2 0 sin k 0 cos k = 2 cos 2 k + 2 sin 2 k = 2. Therefore, A is invertible for every value of k . 6.1: 34 Compute the determinant of A = 4 5 0 0 3 6 0 0 2 7 1 4 1 8 2 3 . det A = det 4 5 3 6 det 1 4 2 3 = (24 - 15)(3 - 8) = - 45. 6.1: 52 For n 4 , define the matrix D n as A n except with (1 , 2) , (2 , 1) entries being 0, and (1 , 3) , (3 , 1) entries being 1. For example, D 4 = 2 0 1 0 0 2 1 0 1 1 2 1 0 0 1 2 Compute det D n in terms of det A n - 1 and det A n - 3 . Find a closed form for det D n . For which n is det D n = 0 ? Denote the ( i, j )-th minor of D n by ( D n ) i,j (the submatrix we get by crossing out the i -th row and j -th columns). Then, det D n = 2 det( D n ) 1 , 1 + 1 det( D n ) 3 , 1 . Observe that ( D n ) 1 , 1 = A n - 1 . ( D n ) 3 , 1 is of the form
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