设 $\boldsymbol{A}$ 为可逆矩阵，令 $\boldsymbol{P}_1=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right), \boldsymbol{P}_2=\left(\begin{array}{lll}1 & 0 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$, 则 $\boldsymbol{A}^{-1} \boldsymbol{P}_1^{100} \boldsymbol{A} \boldsymbol{P}_2^{-1}=(\quad)$.
$\text{A.}$ $\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right)$
$\text{B.}$ $\left(\begin{array}{ccc}1 & 0 & -4 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
$\text{C.}$ $\left(\begin{array}{lll}1 & 0 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
$\text{D.}$ $\left(\begin{array}{ccc}1 & 0 & -4 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right)$