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求矩阵的幂

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其妙 Posted at 2014-6-21 16:18:37 |Read mode
Last edited by hbghlyj at 2025-3-21 05:37:50设矩阵$A=\begin{pmatrix}
1 & 4 &2\\
0 & -3& 4\\
0&4&3
\end{pmatrix} $,求$A^k(k\inN_+)$.

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tommywong Posted at 2014-6-21 16:55:26
$\lambda=1,5,-5$

$\begin{pmatrix} 1 & 4 & 2 \\ 0 & -3 & 4 \\ 0 & 4 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & -2 \\ 0 & 2 & 1 \end{pmatrix} =\begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & -2 \\ 0 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & -5 \end{pmatrix}$

$\begin{pmatrix} 1 & 4 & 2 \\ 0 & -3 & 4 \\ 0 & 4 & 3 \end{pmatrix}^k = \begin{pmatrix} 1 & 2(5^{k-1})+2(-5)^{k-1} & 4(5^{k-1})-(-5)^{k-1}-1 \\ 0 & 5^{k-1}-4(-5)^{k-1} & 2(5^{k-1})+2(-5)^{k-1} \\ 0 & 2(5^{k-1})+2(-5)^{k-1} & 4(5^{k-1})-(-5)^{k-1} \end{pmatrix}$

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 Author| 其妙 Posted at 2014-6-21 18:16:03
回复 2# tommywong
啥子原理哟!

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LLLYSL Posted at 2014-7-1 07:05:54
相似于一个对角阵之后的性质

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2025-4-21 14:18 GMT+8

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