# 分块矩阵的逆运算和次方运算怎么算？

## 一、题目

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

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

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

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

## 二、解析

$$\textcolor{springgreen}{ \left[\begin{array}{ll}A & O \\ O & B\end{array}\right]^{-1}=\left[\begin{array}{cc}A^{-1} & O \\ O & B^{-1}\end{array}\right] }$$

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

$$\textcolor{springgreen}{ \left[\begin{array}{ll}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{B}\end{array}\right]^{n}=\left[\begin{array}{cc}\boldsymbol{A}^{n} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{B}^{n}\end{array}\right] }$$

$$\left[\begin{array}{ll}\boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{O}\end{array}\right]^{2} =$$

$$\left[\begin{array}{ll}\boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{O}\end{array}\right] \left[\begin{array}{ll}\boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{O}\end{array}\right] = \left[\begin{array}{ll}\boldsymbol{AB} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{BA}\end{array}\right]$$

$$\left[\begin{array}{ll}\boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{O}\end{array}\right]^{n} = \left[\begin{array}{ll}\boldsymbol{(AB)^{\frac{n}{2}}} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{(BA)^{\frac{n}{2}}}\end{array}\right]$$

$$\left[\begin{array}{ll}\boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{O}\end{array}\right]^{n} = \left[\begin{array}{ll}\boldsymbol{(AB)^{\frac{n}{2}}A} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{(BA)^{\frac{n}{2}} B}\end{array}\right]$$