# 分块矩阵求逆法：上三角形式（C010）

## 选项

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

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

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

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

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