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Linear Algebra Solutions 64

Linear Algebra Solutions 64 - 2(3 × 2 n 4 n 2 ‚ Hence x...

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or equivalently ˙ x 1 = - 2 x 1 and ˙ y 1 = y 1 . Hence x 1 = x 1 (0) e - 2 t and y 1 = y 1 (0) e t . To determine x 1 (0) and y 1 (0), we note that x 1 (0) y 1 (0) = P - 1 x (0) y (0) = - 1 3 1 - 1 - 5 2 ‚ • 13 22 = 3 7 . Hence x 1 = 3 e - 2 t and y 1 = 7 e t . Consequently x = 2 x 1 + y 1 = 6 e - 2 t + 7 e t and y = 5 x 1 + y 1 = 15 e - 2 t + 7 e t . 4. Introducing the vector X n = x n y n , the system of recurrence relations x n +1 = 3 x n - y n y n +1 = - x n + 3 y n , becomes X n +1 = AX n , where A = 3 - 1 - 1 3 . Hence X n = A n X 0 , where X 0 = 1 2 . To find A n we can use the eigenvalue method. We get A n = 1 2 2 n + 4 n 2 n - 4 n 2 n - 4 n 2 n + 4 n . Hence X n = 1 2 2 n + 4 n 2 n - 4 n 2 n - 4 n 2 n + 4 n ‚ • 1 2 = 1 2 2 n + 4 n + 2(2 n - 4 n ) 2 n - 4 n + 2(2 n + 4 n ) = 1 2 3 × 2 n - 4 n 3 × 2 n + 4 n =
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Unformatted text preview: ) / 2 (3 × 2 n + 4 n ) / 2 ‚ . Hence x n = 1 2 (3 × 2 n-4 n ) and y n = 1 2 (3 × 2 n + 4 n ). 5. Let A = • a b c d ‚ be a real or complex matrix with distinct eigenvalues λ 1 , λ 2 and corresponding eigenvectors X 1 , X 2 . Also let P = [ X 1 | X 2 ]. (a) The system of recurrence relations x n +1 = ax n + by n y n +1 = cx n + dy n 67...
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