22.(10分)(2016•江苏)已知矩阵 $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 0 & -2\end{array}\right]$ ,矩阵 B 的逆矩阵 $\mathrm{B}^{-1}=\left[\begin{array}{ll}1 & -\frac{1}{2} \\ 0 & 2\end{array}\right]$ ,求矩阵AB。
## C.【选修4-4:坐标系与参数方程】
2016_江苏卷 (2016)
22.(10分)(2016•江苏)已知矩阵 $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 0 & -2\end{array}\right]$ ,矩阵 B 的逆矩阵 $\mathrm{B}^{-1}=\left[\begin{array}{ll}1 & -\frac{1}{2} \\ 0 & 2\end{array}\right]$ ,求矩阵AB。
## C.【选修4-4:坐标系与参数方程】
【解答】
(10分)(2016•江苏)已知矩阵 $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 0 & -2\end{array}\right]$ ,矩阵 B 的逆矩阵 $\mathrm{B}^{-1}=\left[\begin{array}{ll}1 & -\frac{1}{2} \\ 0 & 2\end{array}\right]$ ,求矩阵AB。
【分析】依题意,利用矩阵变换求得 $B=\left(B^{-1}\right)^{-1}=\left[\begin{array}{l}\frac{1}{2} \\ \frac{2}{2} \\ \frac{0}{2} \frac{1}{2}\end{array}=\left[\begin{array}{c}1 \frac{1}{4} \\ 0 \frac{1}{2}\end{array}\right]\right.$ ,再利用矩阵乘法的性质可求得答案.
【解答】解:$\because \mathrm{B}^{-1}=\left[\begin{array}{ll}1 & -\frac{1}{2} \\ 0 & 2\end{array}\right]$ ,
$\therefore \mathrm{B}=\left(\mathrm{B}^{-1}\right)^{-1}=\left[\begin{array}{l}\frac{2}{2} \frac{1}{2} \\ \frac{0}{2} \frac{1}{2}\end{array}=\left[\begin{array}{l}1 \frac{1}{4} \\ 0 \frac{1}{2}\end{array}\right], \quad\right.$ 又 $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 0 & -2\end{array}\right]$,
$\therefore \mathrm{AB}=\left[\begin{array}{cc}1 & 2 \\ 0 & -2\end{array}\right]\left[\begin{array}{l}1 \frac{1}{4} \\ 0 \\ \frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}1 & \frac{5}{4} \\ 0 & -1\end{array}\right]$ .
【点评】本题考查逆变换与逆矩阵,考查矩阵乘法的性质,属于中档题.
## C.【选修4-4:坐标系与参数方程】