Question

11．已知矩陣$A=[{\begin{array}{l}1&{\frac{1}{2}}\\ 0&1\end{array}}]，B=[{\begin{array}{l}1&0\\ 0&2\end{array}}]$，設(shè)點$P（{\frac{7}{4}，\frac{5}{2}}）$在矩陣BA對應(yīng)的變換T_BA作用下得到P'點，求點P'的坐標(biāo)．

Answer 1

分析 根據(jù)矩陣的乘法求得矩陣BA，則$[\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}]$，即可求得點P'的坐標(biāo)．

解答 解：由矩陣$A=[{\begin{array}{l}1&{\frac{1}{2}}\\ 0&1\end{array}}]，B=[{\begin{array}{l}1&0\\ 0&2\end{array}}]$，則BA=$[\begin{array}{l}{1}&{0}\\{0}&{2}\end{array}]$$[\begin{array}{l}{1}&{\frac{1}{2}}\\{0}&{1}\end{array}]$=$[\begin{array}{l}{1}&{\frac{1}{2}}\\{0}&{2}\end{array}]$，
從而$[\begin{array}{l}{1}&{\frac{1}{2}}\\{0}&{2}\end{array}]$$[\begin{array}{l}{\frac{7}{4}}\\{\frac{5}{2}}\end{array}]$=$[\begin{array}{l}{3}\\{5}\end{array}]$，
∴點P'的坐標(biāo)（3，5）．

點評 本題考查矩陣的乘法，點的坐標(biāo)變換，考查轉(zhuǎn)化思想，屬于中檔題．