求这个矩阵的 5 次方？可不能真的直接乘 5 次哦

一、题目

已知 $\boldsymbol{A}=\left[\begin{array}{llll}1 & 1 & 0 & 0 \\ 2 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2\end{array}\right]$, 则 $\boldsymbol{A}^{5} = ?$

难度评级：

二、解析

本题用到的基础知识：

1. 《矩阵 n 次幂的三大计算公式》

由题可知：

$$
A^{5}=\left[\begin{array}{llll}1 & 1 & 0 & 0 \\ 2 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2\end{array}\right]^{5}.
$$

若令：

$$
B = \left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right]
$$

$$
C = \left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]
$$

则：

$$
A^{5}=\left[\begin{array}{llll}B^{5} & & 0 & 0 \\ & & 0 & 0 \\ 0 & 0 & & \\ 0 & 0 & & C^{5}\end{array}\right]^{5}.
$$

又：

$$
r\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right] \Rightarrow 1 \Rightarrow
$$

$$
\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right]^{5} = (1+2)^{5-1} \cdot\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right] =
$$

$$
3^{4}\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right] = \left[\begin{array}{cc}3^{4} & 3^{4} \\ 2 \cdot 3^{4} & 2 \cdot 3^{4}\end{array}\right].
$$

且：

$$
{\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]^{5} \Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 2^{5}\end{array}\right]}.
$$

综上可知：

$$
A^{5}=\left[\begin{array}{cccc}3^{4} & 3^{4} & 0 & 0 \\ 2 \cdot 3^{4} & 2 \cdot 3^{4} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2^{5}\end{array}\right]
$$

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