对于抽象矩阵逆矩阵的求解，一定要想方设法引入“矩阵乘法”

一、题目

(A) ${\mathit{A}}^{–1}$ $+$ ${\mathit{B}}^{–1}$

(B) $\mathit{A}(\mathit{A}$ $+$ $\mathit{B}{)}^{–1}\mathit{B}$

(C) $\mathit{A}$ $+$ $\mathit{B}$

(D) $(\mathit{A}+\mathit{B}{)}^{–1}$

难度评级：

二、解析

Note

注意，一般情况下，$(\mathit{A}+\mathit{B}{)}^{–1}$ $\ne$ ${\mathit{A}}^{–1}+{\mathit{B}}^{–1}$

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Tip

求解本题的核心思路就是在矩阵加法中构造出来矩阵乘法，因为只有引入了矩阵乘法，才能方便且正确的使用矩阵逆运算。

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由于 $\mathit{A}$, $\mathit{B}$ 均为可逆矩阵，所以：

$$\mathit{A}{\mathit{A}}^{–1}={\mathit{A}}^{–1}\mathit{A}=\mathit{B}{\mathit{B}}^{–1}={\mathit{B}}^{–1}\mathit{B}=\mathit{E}$$

因此：

$$\begin{array}{r}{\mathit{A}}^{–1}+{\mathit{B}}^{–1}\\ \\ & =\mathit{E}{\mathit{A}}^{–1}+{\mathit{B}}^{–1}\\ \\ & ={\mathit{B}}^{–1}\mathit{B}{\mathit{A}}^{–1}+{\mathit{B}}^{–1}\\ \\ & ={\mathit{B}}^{–1}\mathit{B}{\mathit{A}}^{–1}+{\mathit{B}}^{–1}E\\ \\ & ={\mathit{B}}^{–1}\left(\mathit{B}{\mathit{A}}^{–1}+\mathit{E}\right)\\ \\ & ={\mathit{B}}^{–1}(\mathit{B}{\mathit{A}}^{–1}+\mathit{A}{\mathit{A}}^{–1})\\ \\ & ={\mathit{B}}^{–1}(\mathit{A}+\mathit{B}){\mathit{A}}^{–1}\end{array}$$

进而：

$$\begin{array}{r}{({\mathit{A}}^{–1}+{\mathit{B}}^{–1})}^{–1}\\ \\ & ={\left[{\mathit{B}}^{–1}(\mathit{A}+\mathit{B}){\mathit{A}}^{–1}\right]}^{-1}\\ \\ & =\mathit{A}(\mathit{A}+\mathit{B}{)}^{–1}\mathit{B}\end{array}$$

综上可知，本 题 应 选 B

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