Question

12．關(guān)于x、y的二元線性方程組$\left\{\begin{array}{l}{2x+my=5}\\{nx-3y=2}\end{array}\right.$的增廣矩陣經(jīng)過(guò)變換，最后得到的矩陣為$（\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{1}\end{array}）$，則$\frac{m}{n}$=-$\frac{3}{5}$．

Answer 1

分析 先由矩陣為$（\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{1}\end{array}）$，對(duì)應(yīng)的方程為：$\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$，再由題意得：關(guān)于x、y的二元線性方程組的解為：$\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$，故可求m，n的值，進(jìn)而得到答案．

解答 解：關(guān)于x、y的二元線性方程組$\left\{\begin{array}{l}{2x+my=5}\\{nx-3y=2}\end{array}\right.$的增廣矩陣經(jīng)過(guò)變換可化為：$（\begin{array}{ccc}1&0&3\\ 0&1&1\end{array}\right.）$，
故$\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$是方程組$\left\{\begin{array}{l}{2x+my=5}\\{nx-3y=2}\end{array}\right.$的解，
即$\left\{\begin{array}{l}6+m=5\\ 3n-3=2\end{array}\right.$，
解得：$\left\{\begin{array}{l}m=-1\\ n=\frac{5}{3}\end{array}\right.$
∴$\frac{m}{n}$=-$\frac{3}{5}$，
故答案為：-$\frac{3}{5}$

點(diǎn)評(píng) 本題考查的知識(shí)點(diǎn)是方程組的增廣矩陣，難度不大，屬于基礎(chǔ)題．