Question

10．在平面直角坐標(biāo)系xOy中，先對(duì)曲線C作矩陣A=$[\begin{array}{l}{cosθ}&{-sinθ}\\{sinθ}&{cosθ}\end{array}]$（0＜θ＜2π）所對(duì)應(yīng)的變換，再將所得曲線作矩陣B=$[\begin{array}{l}{1}&{0}\\{0}&{k}\end{array}]$（0＜k＜1）所對(duì)應(yīng)的變換，若連續(xù)實(shí)施兩次變換所對(duì)應(yīng)的矩陣為$[\begin{array}{l}{0}&{-1}\\{\frac{1}{2}}&{0}\end{array}]$，求k，θ的值．

Answer 1

分析 由題意及矩陣乘法的意義可得：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}]$，由矩陣的相等及參數(shù)的范圍即可求解．

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），
∴由題意可得：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}$．

點(diǎn)評(píng) 本題主要考查了矩陣乘法的意義，相等矩陣等知識(shí)的應(yīng)用，屬于基礎(chǔ)題．