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 $\boldsymbol{A}=$
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. $\left(\begin{array}{lll}c & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & a\end{array}\right)$
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 .
高等数学涵盖高等数学基础概念、解题技巧等内容,图文并茂,计算过程清晰严谨。
线性代数以独特的视角解析线性代数,让繁复的知识变得直观明了。
特别专题通过专题的形式对数学知识结构做必要的补充,使所学知识更加连贯坚实。
让考场上没有难做的数学题!