Question

3．已知矩陣A=$|\begin{array}{l}{1}&{a}\\{3}&\end{array}|$，且A$|\begin{array}{l}{19}\\{8}\end{array}|$=$|\begin{array}{l}{3}\\{1}\end{array}|$，求直線l₁：x-y+1=0在矩陣A對(duì)應(yīng)的變換下得到的直線l₂的方程．

Answer 1

分析 根據(jù)矩陣的乘法，$[\begin{array}{l}{1}&{a}\\{3}&\end{array}]$$[\begin{array}{l}{19}\\{8}\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，列方程組即可求得a和b的值，求得矩陣A，P₁（x₁，y₁），P（x，y），$[\begin{array}{l}{1}&{-2}\\{3}&{-7}\end{array}]$$[\begin{array}{l}{{x}_{1}}\\{{y}_{1}}\end{array}]$=$[\begin{array}{l}{x}\\{y}\end{array}]$，根據(jù)矩陣的乘法，列方程求得有$\left\{\begin{array}{l}{{x}_{1}=7x-2y}\\{{y}_{1}=3x-y}\end{array}\right.$，代入x-y+1=0即可得求直線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}&\end{array}]$$[\begin{array}{l}{19}\\{8}\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，
可得：$\left\{\begin{array}{l}{19+8a=3}\\{57+8b=1}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=-2}\\{b=-7}\end{array}\right.$
A=$[\begin{array}{l}{1}&{-2}\\{3}&{-7}\end{array}]$，
設(shè)直線l₁上任一點(diǎn)P₁（x₁，y₁）在矩陣A對(duì)應(yīng)的變換下得到的直線l₂上的對(duì)應(yīng)點(diǎn)P（x，y），
由題意可得$[\begin{array}{l}{1}&{-2}\\{3}&{-7}\end{array}]$$[\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.$，
從而有$\left\{\begin{array}{l}{{x}_{1}=7x-2y}\\{{y}_{1}=3x-y}\end{array}\right.$．
代入方程x-y+1=0得直線l₂的方程4x-y+1=0．（10分）

點(diǎn)評(píng) 本題考查矩陣變換，考查矩二階矩陣的乘法法則，以及求出直線方程利用矩陣的變換所對(duì)應(yīng)的方程，屬于中檔題．