Question

10．在平面直角坐标系xOy中，先对曲线C作矩阵A=$[\begin{array}{l}{cosθ}&{-sinθ}\\{sinθ}&{cosθ}\end{array}]$（0＜θ＜2π）所对应的变换，再将所得曲线作矩阵B=$[\begin{array}{l}{1}&{0}\\{0}&{k}\end{array}]$（0＜k＜1）所对应的变换，若连续实施两次变换所对应的矩阵为$[\begin{array}{l}{0}&{-1}\\{\frac{1}{2}}&{0}\end{array}]$，求k，θ的值．

Answer 1

分析 由题意及矩阵乘法的意义可得：BA=$[\begin{array}{l}{1}&{0}\\{0}&{k}\end{array}]$$[\begin{array}{l}{cosθ}&{-sinθ}\\{sinθ}&{cosθ}\end{array}]$=$[\begin{array}{l}{0}&{-1}\\{\frac{1}{2}}&{0}\end{array}]$，由矩阵的相等及参数的范围即可求解．

解答 解：∵A=$[\begin{array}{l}{cosθ}&{-sinθ}\\{sinθ}&{cosθ}\end{array}]$（0＜θ＜2π），B=$[\begin{array}{l}{1}&{0}\\{0}&{k}\end{array}]$（0＜k＜1），
∴由题意可得：BA=$[\begin{array}{l}{1}&{0}\\{0}&{k}\end{array}]$$[\begin{array}{l}{cosθ}&{-sinθ}\\{sinθ}&{cosθ}\end{array}]$=$[\begin{array}{l}{0}&{-1}\\{\frac{1}{2}}&{0}\end{array}]$，
∴$[\begin{array}{l}{cosθ}&{-sinθ}\\{ksinθ}&{kcosθ}\end{array}]$=$[\begin{array}{l}{0}&{-1}\\{\frac{1}{2}}&{0}\end{array}]$，解得：$\left\{\begin{array}{l}{\stackrel{cosθ=0}{-sinθ=-1}}\\{\stackrel{ksinθ=\frac{1}{2}}{kcosθ=0}}\end{array}\right.$，
∵0＜θ＜2π，0＜k＜1，
∴解得：k=$\frac{1}{2}$，θ=$\frac{π}{2}$．

点评 本题主要考查了矩阵乘法的意义，相等矩阵等知识的应用，属于基础题．