Linear Algebra Solutions 63

Linear Algebra Solutions 63 - 3/5 4/5 . Then we find that...

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Unformatted text preview: 3/5 4/5 . Then we find that the eigenvalues are λ1 = 1 and 2/5 1/5 λ2 = −1/5, with corresponding eigenvectors 2. Let A = 2 1 X1 = and −1 1 X2 = . Then if P = [X1 |X2 ], P is non–singular and P −1 AP = 1 0 0 −1/5 and 1 0 0 −1/5 A=P P −1 . Hence An = P →P = 10 00 1 3 1 3 P −1 P −1 2 −1 1 1 = = 1 0 0 (−1/5)n 1 3 10 00 20 10 11 −1 2 11 −1 2 22 11 = 2/3 2/3 1/3 1/3 . ˙ 3. The given system of differential equations is equivalent to X = AX , where x 3 −2 . and X = A= y 5 −4 21 is a non-singular matrix of eigenvectors corre51 sponding to eigenvalues λ1 = −2 and λ2 = 1. Then The matrix P = P −1 AP = −2 0 01 . The substitution X = P Y , where Y = [x1 , y1 ]t , gives ˙ Y= −2 0 01 66 Y, ...
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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