Question

10．变换T₁是逆时针旋转$\frac{π}{2}$角的旋转变换，对应的变换矩阵是M₁；变换T₂对应的变换矩阵是M₂=$[\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}]$．
（1）点P（2，1）经过变换T₁得到点P′，求P′的坐标；
（2）求曲线y=x²先经过变换T₁，再经过变换T₂所得曲线的方程．

Answer 1

分析 （1）变换T₁对应的变换矩阵M₁=$[\begin{array}{l}{cos\frac{π}{2}}&{-sin\frac{π}{2}}\\{sin\frac{π}{2}}&{cos\frac{π}{2}}\end{array}]$=$[\begin{array}{l}{0}&{-1}\\{1}&{0}\end{array}]$，M₁$[\begin{array}{l}{2}\\{1}\end{array}]$=$[\begin{array}{l}{-1}\\{2}\end{array}]$，即可求得点P在T₁作用下的点P′的坐标；
（2）M=M₂•M₁=$[\begin{array}{l}{1}&{-1}\\{1}&{0}\end{array}]$，由$[\begin{array}{l}{1}&{-1}\\{1}&{0}\end{array}]$$[\begin{array}{l}{{x}_{0}}\\{{y}_{0}}\end{array}]$=$[\begin{array}{l}{x}\\{y}\end{array}]$，求得$\left\{\begin{array}{l}{{x}_{0}=y}\\{{y}_{0}=y-x}\end{array}\right.$，代入y=x²，即可求得经过变换T₂所得曲线的方程．

解答 解：（1）T₁是逆时针旋转$\frac{π}{2}$角的旋转变换，M₁=$[\begin{array}{l}{cos\frac{π}{2}}&{-sin\frac{π}{2}}\\{sin\frac{π}{2}}&{cos\frac{π}{2}}\end{array}]$=$[\begin{array}{l}{0}&{-1}\\{1}&{0}\end{array}]$，
M₁$[\begin{array}{l}{2}\\{1}\end{array}]$=$[\begin{array}{l}{-1}\\{2}\end{array}]$，
所以点P在T₁作用下的点P′的坐标是（-1，2）；
（2）M=M₂•M₁=$[\begin{array}{l}{1}&{-1}\\{1}&{0}\end{array}]$，
设$[\begin{array}{l}{x}\\{y}\end{array}]$是变换后图象上任一点，与之对应的变换前的点是$[\begin{array}{l}{{x}_{0}}\\{{y}_{0}}\end{array}]$，
则M$[\begin{array}{l}{{x}_{0}}\\{{y}_{0}}\end{array}]$=$[\begin{array}{l}{x}\\{y}\end{array}]$，$[\begin{array}{l}{1}&{-1}\\{1}&{0}\end{array}]$$[\begin{array}{l}{{x}_{0}}\\{{y}_{0}}\end{array}]$=$[\begin{array}{l}{x}\\{y}\end{array}]$，
也就是$\left\{\begin{array}{l}{{x}_{0}-{y}_{0}=x}\\{{x}_{0}=y}\end{array}\right.$，即$\left\{\begin{array}{l}{{x}_{0}=y}\\{{y}_{0}=y-x}\end{array}\right.$，
所以所求的曲线方程为y-x=y²．

点评 本题考查矩阵的变换，考查矩阵的乘法，考查点在变换下点的坐标的求法，属于中档题．