# 分块矩阵求逆法：副对角线形式（C010）

## 选项

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

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

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

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

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

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