矩阵乘法和求逆运算都在这道题里了

一、题目

已知 $\boldsymbol{X} \boldsymbol{A}+2 \boldsymbol{E}=\boldsymbol{X}+\boldsymbol{B}$, 其中 $\boldsymbol{A}=\left[\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right], \boldsymbol{B}=\left[\begin{array}{cc}2 & 2 \\ 1 & -1\end{array}\right]$, 则 $\boldsymbol{X}=?$

难度评级：

二、解析

$$
X A+2 E=X+B \Rightarrow
$$

$$
X(A-E)=B-2 E \Rightarrow
$$

$$
x=(B-2 E)(A-E)^{-1}
$$

$$
A-E=\left[\begin{array}{ll}0 & 2 \\ 1 & 0\end{array}\right] \Rightarrow
$$

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

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

$$
B-2 E=\left[\begin{array}{ll}0 & 2 \\ 1 & -3\end{array}\right]
$$

于是：

$$
X = \left[\begin{array}{cc}0 & 2 \\ 1 & -3\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ \frac{1}{2} & 0\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ \frac{-3}{2} & 1\end{array}\right]
$$

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