# 2024年考研数二第08题解析：逆矩阵的计算

## 一、题目

A. $\left(\begin{array}{lll}c & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & b\end{array}\right)$

B. $\left(\begin{array}{lll}b & 0 & 0 \\ 0 & c & 0 \\ 0 & 0 & a\end{array}\right)$

C. $\left(\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right)$

D. $\left(\begin{array}{lll}c & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & a\end{array}\right)$

## 二、解析

$$\boldsymbol{A} = (\boldsymbol{P}^{\top})^{-1} \quad (\boldsymbol{P}^{\top} \boldsymbol{A} \boldsymbol{P}^{2}) \quad (\boldsymbol{P}^{2})^{-1}$$

$$\boldsymbol{P}^{\mathrm{\top}}=\left(\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right) \Rightarrow \left(\boldsymbol{P}^{\mathrm{\top}}\right)^{-1}=\left(\begin{array}{lll}1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$$

$$\boldsymbol{P}^{2}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1\end{array}\right) \Rightarrow \left(\boldsymbol{P}^{2}\right)^{-1}=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{array}\right)$$

\begin{aligned} \boldsymbol{A} & =\left(\boldsymbol{P}^{\mathrm{\top}}\right)^{-1}\left(\begin{array}{ccc}a+2 c & 0 & c \\ 0 & b & 0 \\ 2 c & 0 & c\end{array}\right)\left(\boldsymbol{P}^{2}\right)^{-1} \\ \\ & =\left(\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{ccc}a+2 c & 0 & c \\ 0 & b & 0 \\ 2 c & 0 & c\end{array}\right)\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{array}\right) \\ \\ & =\left(\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right) \end{aligned}