2024年考研数二第08题解析：逆矩阵的计算

一、题目

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

难度评级：

二、解析

由于：

$$\mathit{A}=({\mathit{P}}^{\mathrm{\top }}{)}^{-1}({\mathit{P}}^{\mathrm{\top }}\mathit{A}{\mathit{P}}^2)({\mathit{P}}^2{)}^{-1}$$

且，由求逆变换法,可得：

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

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

于是：

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

综上可知，本题应选 C .

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