M 340L HW 10 Solutions

# Thus x0 c1 v1 c2 v2 but v1 v2 are eigenvectors

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Unformatted text preview: nd x0 = 3v1 + 4v2 . 1 2k 6. xk = 6(2k )v1 + 2 v2 1 2k v2 Explanation: Since v1 , v2 are eigenvectors corresponding to distinct eigenvalues of A, they form an eigenbasis for R2 . Thus x0 = c1 v1 + c2 v2 . But v1 , v2 are eigenvectors corresponding to 1 respective eigenvalues 2, 2 so To compute c1 , c2 we apply row reduction to the augmented matrix x1 = Ax0 = 3Av1 + 4Av2 −1 1 +2 = 6v1 + 2v2 = 6 1 1 [ v1 v2 x0 ] = . Consequently, 1 5 ∼ 0 1 1 −1 11 03 , 1 −2 for then c1 = 3, c2 = −2 and x0 = 3v1 − 2v2 . x1 = 4 8 But v1 , v2 are eigenvectors corresponding to 1 respective eigenvalues 2, 2 so set . . 1 2k xk = 3(2k )v1 − 2 003 10.0 points When A is a 2 × 2 matrix with eigenvalues 1 2, 2 and corresponding eigenvectors 1 , v1 = 1 −1 , v2 = 1 determine the solution {xk } of the diﬀerence equation xk+1 = Axk , x0 = 5 . 1 v2 for k ≥ 0. Then x0 is the given initial value and Axk = 3 2k Av1 − 2 = 3 2k+1 v1 − 2 1 2k Av2 1 2k+1 v2 = xk+1 , so {xk } solve the diﬀerence equation. 004 10.0 points cruz (fmc326)...
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## This document was uploaded on 03/16/2014 for the course M 340L at University of Texas.

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