设 $\boldsymbol{A}$ 为三阶矩阵, $\boldsymbol{P}=\left(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3\right)$ 为可逆矩阵,使得 $\boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2\end{array}\right)$, 则 $\boldsymbol{A}^2\left(\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2+\right.$ $\boldsymbol{\alpha}_3$ ) 是
$ \text{A.} $ $\boldsymbol{\alpha}_1+2 \boldsymbol{\alpha}_3$ $ \text{B.} $ $\boldsymbol{\alpha}_2+2 \boldsymbol{\alpha}_3$ $ \text{C.} $ $\boldsymbol{\alpha}_1+4 \boldsymbol{\alpha}_3$ $ \text{D.} $ $\boldsymbol{\alpha}_2+4 \boldsymbol{\alpha}_3$
【答案】 C

【解析】 【解】
$$
\begin{aligned}
\text { 由 } \boldsymbol{P}^{-1} \boldsymbol{A P}= & \left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 2
\end{array}\right) \text { 得 } \boldsymbol{P}^{-1} \boldsymbol{A}^2 \boldsymbol{P}=\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 4
\end{array}\right), \boldsymbol{A}^2 \boldsymbol{P}=\boldsymbol{P}\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 4
\end{array}\right), \\
& \boldsymbol{A}^2\left(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3\right)=\left(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3\right)\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 4
\end{array}\right)=\left(\boldsymbol{\alpha}_1, \mathbf{0}, 4 \boldsymbol{\alpha}_3\right),
\end{aligned}
$$

$$
\boldsymbol{A}^2 \boldsymbol{\alpha}_1=\boldsymbol{\alpha}_1, \boldsymbol{A}^2 \boldsymbol{\alpha}_2=\mathbf{0}, \boldsymbol{A}^2 \boldsymbol{\alpha}_3=4 \boldsymbol{\alpha}_3 .
$$
所以 $\boldsymbol{A}^2\left(\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2+\boldsymbol{\alpha}_3\right)=\boldsymbol{\alpha}_1+4 \boldsymbol{\alpha}_3$. 故选 C.
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