$\text{A.}$ $\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{array}\right)$. $\text{B.}$ $\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 6 \\ 0 & 0 & 3\end{array}\right)$. $\text{C.}$ $\left(\begin{array}{lll}0 & 1 & 0 \\ 2 & 0 & 0 \\ 0 & 0 & 3\end{array}\right)$. $\text{D.}$ $\left(\begin{array}{lll}0 & 1 & 0 \\ 2 & 0 & 6 \\ 0 & 0 & 3\end{array}\right)$.
【答案】 D

【解析】 设初等矩阵 $\boldsymbol{P}=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 1\end{array}\right), \boldsymbol{Q}=\left(\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)$, 则 $\boldsymbol{A}_1=\boldsymbol{P A}, \boldsymbol{B}_1=\boldsymbol{B} \boldsymbol{Q}$.
\begin{aligned} &\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{3} \end{array}\right)\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{array}\right) \boldsymbol{A}_1 \boldsymbol{B}_1=\boldsymbol{E} \text {, 则 } \boldsymbol{A}_1 \boldsymbol{B}_1=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) \text {, 又 } \boldsymbol{A}_1 \boldsymbol{B}_1=\boldsymbol{P A B Q} \text {, 则 } \\ &\boldsymbol{A} \boldsymbol{B}=\boldsymbol{P}^{-1}\left(\boldsymbol{A}_1 \boldsymbol{B}_1\right) \boldsymbol{Q}^{-1}=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right)\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right)\left(\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right)=\left(\begin{array}{lll} 0 & 1 & 0 \\ 2 & 0 & 6 \\ 0 & 0 & 3 \end{array}\right) . \\ & \end{aligned}