Question

3．設(shè)2階方矩陣A=$（\begin{array}{l}{a}&\\{c}&u1kohse\end{array}）$，則矩陣A所對(duì)應(yīng)的矩陣變換為：$（\begin{array}{l}{x}\\{y}\end{array}）$=$（\begin{array}{l}{a}&\\{c}&zfr6vp4\end{array}）$$（\begin{array}{l}{x′}\\{y′}\end{array}）$，其意義是把點(diǎn)P（x，y）變換為點(diǎn)Q（x′，y′），矩陣A叫做變換矩陣．
（1）當(dāng)變換矩陣A₁=$（\begin{array}{l}{1}&{2}\\{2}&{1}\end{array}）$時(shí)，點(diǎn)P₁（-1，1），P₂（-3，1）經(jīng)矩陣變換后得到點(diǎn)分別是Q₁，Q₂，求過點(diǎn)Q₁，Q₂的直線的點(diǎn)向式方程．
（2）當(dāng)變換矩陣A₂=$（\begin{array}{l}{1}&{3}\\{8}&{-1}\end{array}）$時(shí)，若直線上的任意點(diǎn)P（x，y）經(jīng)矩陣變換后得到的點(diǎn)Q仍在該直線上，求直線方程．

Answer 1

分析 （1）$（{\begin{array}{l}{-1}\\ 1\end{array}}）=（{\begin{array}{l}1&2\\ 2&1\end{array}}）（{\begin{array}{l}{x^'}\\{{y^'}}\end{array}}）=（{\begin{array}{l}{{x^'}+2{y^'}}\\{2{x^'}+{y^'}}\end{array}}）$，由$\left\{{\begin{array}{l}{{x^'}+2{y^'}=-1}\\{2{x^'}+{y^'}=1}\end{array}⇒\left\{{\begin{array}{l}{{x^'}=1}\\{{y^'}=-1}\end{array}}\right.}\right.$，求得Q₁點(diǎn)坐標(biāo)，根據(jù)矩陣的變換，求得Q₂坐標(biāo)，求得$\overrightarrow{{Q}_{1}{Q}_{2}}$=（$\frac{2}{3}$，-$\frac{4}{3}$），根據(jù)點(diǎn)方向式，求得過點(diǎn)Q₁，Q₂的直線的點(diǎn)方向式方程；
（2）根據(jù)矩陣的坐標(biāo)變換，求得$\left\{{\begin{array}{l}{x={x^'}+3{y^'}}\\{y=8{x^'}-{y^'}}\end{array}}\right.⇒\left\{{\begin{array}{l}{{x^'}=\frac{x+3y}{25}}\\{{y^'}=\frac{8x-y}{25}}\end{array}}\right.$，將x′和y′代入直線l₁方程，由l₁與l₂重合，求得a和b的關(guān)系，由D_x=0，求得c的值，D_y=0，求得直線方程．

解答 解：（1）$（{\begin{array}{l}{-1}\\ 1\end{array}}）=（{\begin{array}{l}1&2\\ 2&1\end{array}}）（{\begin{array}{l}{x^'}\\{{y^'}}\end{array}}）=（{\begin{array}{l}{{x^'}+2{y^'}}\\{2{x^'}+{y^'}}\end{array}}）$，
則$\left\{{\begin{array}{l}{{x^'}+2{y^'}=-1}\\{2{x^'}+{y^'}=1}\end{array}⇒\left\{{\begin{array}{l}{{x^'}=1}\\{{y^'}=-1}\end{array}}\right.}\right.$，
∴點(diǎn)Q₁（1，-1）．
同理點(diǎn)${Q_2}（\frac{5}{3}，-\frac{7}{3}）$．$\overrightarrow{{Q}_{1}{Q}_{2}}$=（$\frac{2}{3}$，-$\frac{4}{3}$），
直線Q₁Q₂的點(diǎn)向式為 $\frac{x-1}{{\frac{2}{3}}}=\frac{y+1}{{-\frac{4}{3}}}$，即$\frac{x-1}{1}=\frac{y+1}{-2}$．
（2）$（{\begin{array}{l}x\\ y\end{array}}）=（{\begin{array}{l}1&3\\ 8&{-1}\end{array}}）（{\begin{array}{l}{x^'}\\{{y^'}}\end{array}}）=（{\begin{array}{l}{{x^'}+3{y^'}}\\{8{x^'}-{y^'}}\end{array}}）$，
$\left\{{\begin{array}{l}{x={x^'}+3{y^'}}\\{y=8{x^'}-{y^'}}\end{array}}\right.⇒\left\{{\begin{array}{l}{{x^'}=\frac{x+3y}{25}}\\{{y^'}=\frac{8x-y}{25}}\end{array}}\right.$．
設(shè)l₁：ax+by+c=0（a，b不全為0）${l_2}：a\frac{x+3y}{25}+b\frac{8x-y}{25}+c=0$，
即  （a+8b）x+（3a-b）y+25c=0
由題知l₁與l₂重合得$D=|{\begin{array}{l}a&b\\{a+8b}&{3a-b}\end{array}}|=3{a^2}-2ab-8{b^2}=0$，
∴a=2b或$a=-\frac{4}{3}b$，
${D_x}=|{\begin{array}{l}{-c}&b\\{-25c}&{3a-b}\end{array}}|=0$，
得  c=0，D_y=$|\begin{array}{l}{a}&{-c}\\{a+8b}&{-25c}\end{array}|$=0，
∴2bx+by=0或   $（-\frac{4}{3}b）x+by=0$，
即  2x+y=0或4x-3y=0．

點(diǎn)評(píng) 本題考查矩陣變換問題，考查矩陣的求法，考查運(yùn)算能力與轉(zhuǎn)換思想，屬于中檔題．