设 $\boldsymbol{A}$ 为 3 阶矩阵, $\boldsymbol{P}$ 为 3 阶可逆矩阵, 且 $\boldsymbol{P}^{-1} \boldsymbol{A P}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{array}\right)$. 若 $\boldsymbol{P}=\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}\right)$, $Q=\left(\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}\right)$, 则 $Q^{-1} A Q=(\quad)$
$\text{A.}$ $\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{array}\right)$.
$\text{B.}$ $\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{array}\right)$.
$\text{C.}$ $\left(\begin{array}{lll}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{array}\right)$.
$\text{D.}$ $\left(\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{array}\right)$.