# 分块矩阵求逆法：主对角线形式（C010）

## 选项

[A].   $\left(\begin{array}{ll}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{B}\end{array}\right)^{-1}$ $=$ $\left(\begin{array}{ll}\boldsymbol{B}^{-1} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{A}^{-1}\end{array}\right)$

[B].   $\left(\begin{array}{ll}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{B}\end{array}\right)^{-1}$ $=$ $\left(\begin{array}{ll}-\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & -\boldsymbol{B}\end{array}\right)$

[C].   $\left(\begin{array}{ll}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{B}\end{array}\right)^{-1}$ $=$ $\left(\begin{array}{ll}\boldsymbol{A}^{-1} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{B}^{-1}\end{array}\right)$

[D].   $\left(\begin{array}{ll}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{B}\end{array}\right)^{-1}$ $=$ $\left(\begin{array}{ll}\boldsymbol{O} & \boldsymbol{A}^{-1} \\ \boldsymbol{B}^{-1} & \boldsymbol{O}\end{array}\right)$

$\left(\begin{array}{ll}\boldsymbol{\textcolor{orange}{A}} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{\textcolor{cyan}{B}}\end{array}\right)^{\textcolor{red}{-1}}$ $=$ $\left(\begin{array}{ll}\boldsymbol{\textcolor{orange}{A}}^{\textcolor{red}{-1}} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{\textcolor{cyan}{B}}^{\textcolor{red}{-1}}\end{array}\right)$