Question

12．已知矩陣A的逆矩陣A^-1=$[\begin{array}{l}{\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\\{-\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\end{array}]$．求曲線xy=1在矩陣A所對(duì)應(yīng)的變換作用下所得的曲線方程．

Answer 1

分析 根據(jù)矩陣變換的特點(diǎn)代入計(jì)算即可．

解答 解：設(shè)xy=1上任意一點(diǎn)（x，y）在矩陣A所對(duì)應(yīng)的變換作用下對(duì)應(yīng)的點(diǎn)（x′，y′），
則$[\begin{array}{l}{x}\\{y}\end{array}]$=A^-1$[\begin{array}{l}{x′}\\{y′}\end{array}]$=$[\begin{array}{l}{\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\\{-\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\end{array}]$$[\begin{array}{l}{x′}\\{y′}\end{array}]$，
由此得$\left\{\begin{array}{l}{x=\frac{\sqrt{2}}{2}（x′+y′）}\\{y=\frac{\sqrt{2}}{2}（y′-x′）}\end{array}\right.$，
代入方程xy=1，得y′²-x′²=2．
所以xy=1在矩陣A所對(duì)應(yīng)的線性變換作用下的曲線方程為y²-x²=2．

點(diǎn)評(píng) 本題考查矩陣的變換等知識(shí)，注意解題方法的積累，屬于基礎(chǔ)題．