2024年考研数二第08题解析：逆矩阵的计算

一、题目

设 $\boldsymbol{A}$ 为三阶矩阵, $\boldsymbol{P}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{array}\right)$, 若 $\boldsymbol{P}^{\mathrm{\top}} \boldsymbol{A} \boldsymbol{P}^{2}=\left(\begin{array}{ccc}a+2 c & 0 & c \\ 0 & b & 0 \\ 2 c & 0 & c\end{array}\right)$, 则 $\boldsymbol{A}=$

难度评级：

二、解析

由于：

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

且，由求逆变换法,可得：

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

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

于是：

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

综上可知，本题应选 C .

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