矩阵乘法一般是不能交换的：除非他们相乘得单位矩阵

一、题目

难度评级：

二、解析

首先，由 $\boldsymbol{BA}$ $=$ $\boldsymbol{E}$ 可得：

$$
\boldsymbol{B} ^{-1} = \boldsymbol{A}
$$

于是：

$$
\boldsymbol{AB} = \boldsymbol{B} ^{-1} \boldsymbol{B} = E
$$

又因为：

$$
\textcolor{magenta}{\boldsymbol{A} ^{\top} \boldsymbol{B} ^{\top}} = (\boldsymbol{AB}) ^{\top} = \boldsymbol{E} ^{\top} = \textcolor{magenta}{\boldsymbol{E}}
$$

所以：

$$
\begin{aligned}
& \textcolor{springgreen}{\boldsymbol{ \boldsymbol{B} \left[ \boldsymbol{E} + \boldsymbol{A} \left( \boldsymbol{E} + 2 \boldsymbol{B} ^{\top} \boldsymbol{A} ^{\top} \right) ^{-1} \boldsymbol{B} \right] \boldsymbol{A} }} \\ \\
\Rightarrow & \boldsymbol{B} \left[ \boldsymbol{E} + \boldsymbol{A} \left( \boldsymbol{E} + 2 \textcolor{magenta}{\boldsymbol{B} ^{\top} \boldsymbol{A} ^{\top}} \right) ^{-1} \boldsymbol{B} \right] \boldsymbol{A} \\ \\
\Rightarrow & \boldsymbol{B} \left[ \boldsymbol{E} + \boldsymbol{A} \left( \boldsymbol{E} + 2 \textcolor{magenta}{E} \right) ^{-1} \boldsymbol{B} \right] \boldsymbol{A} \\ \\
\Rightarrow & \boldsymbol{B} \left[ \boldsymbol{E} + \boldsymbol{A} \textcolor{pink}{\left( 3 E \right) ^{-1} } \boldsymbol{B} \right] \boldsymbol{A} \\ \\
\Rightarrow & \boldsymbol{B} \left[ \boldsymbol{E} + \boldsymbol{A} \textcolor{pink}{\frac{1}{3} E} \boldsymbol{B} \right] \boldsymbol{A} \\ \\
\Rightarrow & \textcolor{orangered}{\boldsymbol{B}} \left[ \boldsymbol{E} + \frac{1}{3} \boldsymbol{A} \boldsymbol{B} \right] \textcolor{brown}{\boldsymbol{A}} \\ \\
\Rightarrow & \textcolor{orangered}{\boldsymbol{B}} \boldsymbol{E} \textcolor{brown}{\boldsymbol{A}} + \frac{1}{3} (\textcolor{orangered}{\boldsymbol{B}} \boldsymbol{A}) (\boldsymbol{B} \textcolor{brown}{\boldsymbol{A}}) \\ \\
\Rightarrow & \textcolor{yellow}{\boldsymbol{BA}} + \frac{1}{3} \textcolor{#1e90ff}{\boldsymbol{E E}} \\ \\
\Rightarrow & \textcolor{yellow}{\boldsymbol{E}} + \frac{1}{3} \textcolor{#1e90ff}{\boldsymbol{E}} \\ \\
\Rightarrow & \textcolor{springgreen}{\boldsymbol{ \frac{4}{3} \boldsymbol{E} }}
\end{aligned}
$$

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