分块矩阵的逆运算和次方运算怎么算？

一、题目

已知 $A, B$ 均是 $n$ 阶可逆矩阵, 则以下错误的是哪个？

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

难度评级：

二、解析

注意：本提问的是“错误的”是哪个，而不是“正确的”是哪个。

根据主对角线分块矩阵的逆运算法则和副对角线分块矩阵的逆运算法则，A、B、C 选项都是原有的公式，所以一定是正确的：

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

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

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

对于 D 选项，我们只需要令 $n = 2$ 计算一下即可判断：

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

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

事实上，当 $n$ 为偶数时：

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

当 $n$ 为奇数时：

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

综上可知，正确选项为 D.

相关例题

1. 2023年考研数二第08题解析：伴随矩阵的性质在分块矩阵上的推广

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