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

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：

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

方法二：利用矩阵的性质先变形

已知：

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

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

于是：

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

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

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

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

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

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

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