Question

3．已知矩阵A=$|\begin{array}{l}{1}&{a}\\{3}&{b}\end{array}|$，且A$|\begin{array}{l}{19}\\{8}\end{array}|$=$|\begin{array}{l}{3}\\{1}\end{array}|$，求直线l₁：x-y+1=0在矩阵A对应的变换下得到的直线l₂的方程．

Answer 1

分析 根据矩阵的乘法，$[\begin{array}{l}{1}&{a}\\{3}&{b}\end{array}]$$[\begin{array}{l}{19}\\{8}\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，列方程组即可求得a和b的值，求得矩阵A，P₁（x₁，y₁），P（x，y），$[\begin{array}{l}{1}&{-2}\\{3}&{-7}\end{array}]$$[\begin{array}{l}{{x}_{1}}\\{{y}_{1}}\end{array}]$=$[\begin{array}{l}{x}\\{y}\end{array}]$，根据矩阵的乘法，列方程求得有$\left\{\begin{array}{l}{{x}_{1}=7x-2y}\\{{y}_{1}=3x-y}\end{array}\right.$，代入x-y+1=0即可得求直线l₂的方程．

解答 解：∵A$[\begin{array}{l}{19}\\{8}\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，即$[\begin{array}{l}{1}&{a}\\{3}&{b}\end{array}]$$[\begin{array}{l}{19}\\{8}\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，
可得：$\left\{\begin{array}{l}{19+8a=3}\\{57+8b=1}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=-2}\\{b=-7}\end{array}\right.$
A=$[\begin{array}{l}{1}&{-2}\\{3}&{-7}\end{array}]$，
设直线l₁上任一点P₁（x₁，y₁）在矩阵A对应的变换下得到的直线l₂上的对应点P（x，y），
由题意可得$[\begin{array}{l}{1}&{-2}\\{3}&{-7}\end{array}]$$[\begin{array}{l}{{x}_{1}}\\{{y}_{1}}\end{array}]$=$[\begin{array}{l}{x}\\{y}\end{array}]$，
所以$\left\{\begin{array}{l}{{x}_{1}-2{y}_{1}=x}\\{3{x}_{2}-7{y}_{1}=y}\end{array}\right.$，
从而有$\left\{\begin{array}{l}{{x}_{1}=7x-2y}\\{{y}_{1}=3x-y}\end{array}\right.$．
代入方程x-y+1=0得直线l₂的方程4x-y+1=0．（10分）

点评 本题考查矩阵变换，考查矩二阶矩阵的乘法法则，以及求出直线方程利用矩阵的变换所对应的方程，属于中档题．