已知 $\boldsymbol{A}, \boldsymbol{B}$ 均为 3 阶矩阵, 将 $\boldsymbol{A}$ 的第 3 行的 (-2) 倍加至第 2 行得 $\boldsymbol{A}_1$, 将 $\boldsymbol{B}$ 的第 1,2两列互换得 $\boldsymbol{B}_1$, 再将乘积矩阵 $\boldsymbol{A}_1 \boldsymbol{B}_1$ 的第 2 行乘以 $\frac{1}{2}$, 第 3 行乘以 $\frac{1}{3}$ 得单位矩阵 $\boldsymbol{E}$, 则 $\boldsymbol{A B}=$
$ \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 $\boldsymbol{A}_1=\boldsymbol{P A}, \boldsymbol{B}_1=\boldsymbol{B} \boldsymbol{Q}$.
$$
\begin{aligned}
&\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \frac{1}{3}
\end{array}\right)\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & \frac{1}{2} & 0 \\
0 & 0 & 1
\end{array}\right) \boldsymbol{A}_1 \boldsymbol{B}_1=\boldsymbol{E} \text {, 则 } \boldsymbol{A}_1 \boldsymbol{B}_1=\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}\right) \text {, 又 } \boldsymbol{A}_1 \boldsymbol{B}_1=\boldsymbol{P A B Q} \text {, 则 } \\
&\boldsymbol{A} \boldsymbol{B}=\boldsymbol{P}^{-1}\left(\boldsymbol{A}_1 \boldsymbol{B}_1\right) \boldsymbol{Q}^{-1}=\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 2 \\
0 & 0 & 1
\end{array}\right)\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}\right)\left(\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right)=\left(\begin{array}{lll}
0 & 1 & 0 \\
2 & 0 & 6 \\
0 & 0 & 3
\end{array}\right) . \\
&
\end{aligned}
$$
故应选 (D).
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