Question

11．已知矩阵$A=[{\begin{array}{l}1&{\frac{1}{2}}\\ 0&1\end{array}}]，B=[{\begin{array}{l}1&0\\ 0&2\end{array}}]$，设点$P（{\frac{7}{4}，\frac{5}{2}}）$在矩阵BA对应的变换T_BA作用下得到P'点，求点P'的坐标．

Answer 1

分析 根据矩阵的乘法求得矩阵BA，则$[\begin{array}{l}{1}&{\frac{1}{2}}\\{0}&{2}\end{array}]$$[\begin{array}{l}{\frac{7}{4}}\\{\frac{5}{2}}\end{array}]$=$[\begin{array}{l}{3}\\{5}\end{array}]$，即可求得点P'的坐标．

解答 解：由矩阵$A=[{\begin{array}{l}1&{\frac{1}{2}}\\ 0&1\end{array}}]，B=[{\begin{array}{l}1&0\\ 0&2\end{array}}]$，则BA=$[\begin{array}{l}{1}&{0}\\{0}&{2}\end{array}]$$[\begin{array}{l}{1}&{\frac{1}{2}}\\{0}&{1}\end{array}]$=$[\begin{array}{l}{1}&{\frac{1}{2}}\\{0}&{2}\end{array}]$，
从而$[\begin{array}{l}{1}&{\frac{1}{2}}\\{0}&{2}\end{array}]$$[\begin{array}{l}{\frac{7}{4}}\\{\frac{5}{2}}\end{array}]$=$[\begin{array}{l}{3}\\{5}\end{array}]$，
∴点P'的坐标（3，5）．

点评 本题考查矩阵的乘法，点的坐标变换，考查转化思想，属于中档题．