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Unformatted text preview: Solutions for Quiz on March 29 We need to diagonalize the matrix, A = 1 1 1 1 1 . First we find the eigenvalues. We need to solve fl fl fl fl 1 λ 1 λ 1 1 1 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 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  1 1 ....
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This note was uploaded on 01/12/2010 for the course MATH 421 taught by Professor Nataliakomarova during the Winter '07 term at San Jose State University .
 Winter '07
 nataliakomarova
 Determinant, Factors

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