Question

1．關(guān)于x、y的方程組$\left\{\begin{array}{l}{2x+my=5}\\{nx-4y=2}\end{array}\right.$的增廣矩陣經(jīng)過變換后得到$（\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{1}\end{array}）$，則$（\begin{array}{l}{m}\\{n}\end{array}）$=$（\begin{array}{l}{-1}\\{2}\end{array}）$．

Answer 1

分析 由題意可知矩陣為$（\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{1}\end{array}）$，對應(yīng)的方程組為：$\left\{\begin{array}{l}{1•x+0•y=3}\\{0•x+1•y=1}\end{array}\right.$，則$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$，代入方程組，即可求得m和n的值，即可求得矩陣$（\begin{array}{l}{m}\\{n}\end{array}）$的值．

解答 解：矩陣為$（\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{1}\end{array}）$，對應(yīng)的方程組為：$\left\{\begin{array}{l}{1•x+0•y=3}\\{0•x+1•y=1}\end{array}\right.$，解得：$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$，
由題意得：關(guān)于x、y的二元線性方程組$\left\{\begin{array}{l}{2x+my=5}\\{nx-4y=2}\end{array}\right.$的解為：$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$，
∴$\left\{\begin{array}{l}{2×3+m=5}\\{3n-4×1=2}\end{array}\right.$，解得：$\left\{\begin{array}{l}{m=-1}\\{n=2}\end{array}\right.$，
$（\begin{array}{l}{m}\\{n}\end{array}）$=$（\begin{array}{l}{-1}\\{2}\end{array}）$，
故答案為：$（\begin{array}{l}{-1}\\{2}\end{array}）$．

點評 本題的考點是二元一次方程組的矩陣形式，主要考查了幾種特殊的矩陣變換，解答的關(guān)鍵是對增廣矩陣的理解，利用方程組同解解決問題，屬于基礎(chǔ)題．