不是所有题目都有巧妙做法：这道常数矩阵的逆矩阵题目直接算就很简单

一、题目

已知 $\mathit{A}=\left[\begin{array}{lll}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]\left[\begin{array}{lll}1& 0& 0\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]$, 则 ${\left(\frac{1}{3}\mathit{A}\right)}^{-1}=?$

难度评级：

二、解析

方法一：直接算

$$A=\left[\begin{array}{lll}0& 0& 1\\ 0& 3& 2\\ 1& 0& 0\end{array}\right]\Rightarrow$$

$$\left[\begin{array}{llllll}0& 0& 1& 1& 0& 0\\ 0& 3& 2& 0& 1& 0\\ 1& 0& 0& 0& 0& 1\end{array}\right]\Rightarrow$$

$$\left[\begin{array}{llllll}1& 0& 0& 0& 0& 1\\ 0& 1& 0& \frac{-2}{3}& \frac{1}{3}& 0\\ 0& 0& 1& 1& 0& 0\end{array}\right].$$

于是：

$$A^{-1}=\left[\begin{array}{ccc}0& 0& 1\\ \frac{-2}{3}& \frac{1}{3}& 0\\ 1& 0& 0\end{array}\right]$$

进而：

$${\left(\frac{1}{3}A\right)}^{-1}=3A^{-1}=3\left[\begin{array}{ccc}0& 0& 1\\ \frac{-2}{3}& \frac{1}{3}& 0\\ 1& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 3\\ -2& 1& 0\\ 3& 0& 0\end{array}\right].$$

方法二：利用矩阵的性质先变形

已知：

$$(kA{)}^{-1}=\frac{1}{k}{A}^{-1}$$

$$(ABC{)}^{-1}=C^{-1}{B}^{-1}{A}^{-1}$$

于是：

$${\left(\frac{1}{3}A\right)}^{-1}=3A^{-1}=$$

$$3{\left[\begin{array}{lll}1& 0& 0\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]}^{-1}{\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]}^{-1}{\left[\begin{array}{lll}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]}^{-1}=$$

$$3\left[\begin{array}{lll}1& 0& 0\\ 0& \frac{1}{3}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& -2\\ 0& 0& 1\end{array}\right]\left[\begin{array}{lll}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]\Rightarrow$$

$$3A^{-1}=3\left[\begin{array}{ccc}1& 0& 0\\ 0& \frac{1}{3}& \frac{-2}{3}\\ 0& 0& 1\end{array}\right]\left[\begin{array}{lll}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]=$$

$$3\left[\begin{array}{ccc}0& 0& 1\\ \frac{-2}{3}& \frac{1}{3}& 0\\ 1& 0& 0\end{array}\right]\Rightarrow$$

$$3A^{-1}=\left[\begin{array}{ccc}0& 0& 3\\ -2& 1& 0\\ 3& 0& 0\end{array}\right].$$

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