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 $\boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P}=\left(\begin{array}{llll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2\end{array}\right)$, 则 $\boldsymbol{A}^2\left(\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2+\right.$ $\left.\boldsymbol{\alpha}_3\right)$ 是
$\text{A.}$ $\boldsymbol{\alpha}_1+2 \boldsymbol{\alpha}_3$
$\text{B.}$ $\boldsymbol{\alpha}_2+2 \boldsymbol{\alpha}_3$
$\text{C.}$ $\boldsymbol{\alpha}_1+4 \boldsymbol{\alpha}_3$
$\text{D.}$ $\boldsymbol{\alpha}_2+4 \boldsymbol{\alpha}_3$