Re: [中學] 長得有點像特徵型式的矩陣
: 想過用特徵函數去反推
: 不過因為是2*2不是1*1 腦袋就打結了@@
det(A - λI) = 0 => λ = 2, 5
A = Q^(-1) D[2,5] Q
Q你應該要會求
A^(n+1) - 2A^n = α[A - 2I]
=> D[2^(n+1) - 2*2^n, 5^(n+1) - 2*5^n] = αD[0, 3]
=> α = 5^n
A^(n+1) - 5A^n = β[A - 5I]
=> D[2^(n+1) - 5*2^n, 5^(n+1) - 5*5^n] = βD[-3, 0]
=> β = 2^n
A^n = Q^(-1)D[2^n, 5^n]Q
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補完cuttlefish板友的做法
(A - 5I)(A - 2I) = 0
=> A(A - 2I) = 5(A - 2I)
所以A^(n + 1) - 2A^n = A^n (A - 2I)
= A^(n - 1) [5(A - 2I)] = ...
= 5^n (A - 2I) => α = 5^n
同理A^(n + 1) - 5A^n = A^n (A - 5I)
= A^(n - 1) 2(A - 5I)
= 2^n (A - 5I) => β = 2^n
第一式 - 第二式：
3A^n = [5^n - 2^n]A + [-2 * 5^n + 5 * 2^n]I
=> A^n = {[5^n - 2^n] / 3}A + {[-2 * 5^n + 5 * 2^n] / 3}I
感謝cuttlefish板友的作法
※ 編輯: Honor1984 (61.56.10.112), 12/27/2016 16:47:09
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