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quiz9 - Solutions for Quiz on March 29 We need to...

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Solutions for Quiz on March 29 We need to diagonalize the matrix, A = 1 0 0 - 1 0 1 1 1 0 . First we find the eigenvalues. We need to solve fl fl fl fl 1 - λ 0 0 - 1 0 - λ 1 1 1 0 - λ fl fl fl fl = 0 . We expand the determinant in cofactors of the first row. This row has only one nonzero element, so we have (1 - λ )( - 1) 1+1 M 13 = (1 - λ ) fl fl fl fl - λ 1 1 - λ fl fl fl fl = 0 . The minor is given by λ 2 - 1, so the cherecteristic equation becomes (1 - λ )( λ 2 - 1) = 0 . This has roots λ 1 = 1, λ 2 = 1 and λ 3 = - 1. Now we need to find the eigenvectors. We start from the eigenvector corresponding to λ 3 . We have, 2 0 0 - 1 1 1 1 1 1 x 1 x 2 x 3 = 0 . From the first equation, we have 2 x 1 = 0, which means x 1 = 0. From the second and third equations, we have, 0+ x 2 + x 3 = 0, which means x 2 = - x 3 . Therefore, the corresponding solution is α 0 - 1 1 . 1
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Remark. This equation can also be solved by row reduction, if you like.
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