Question

1．关于x、y的方程组$\left\{\begin{array}{l}{2x+my=5}\\{nx-4y=2}\end{array}\right.$的增广矩阵经过变换后得到$（\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{1}\end{array}）$，则$（\begin{array}{l}{m}\\{n}\end{array}）$=$（\begin{array}{l}{-1}\\{2}\end{array}）$．

Answer 1

分析 由题意可知矩阵为$（\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{1}\end{array}）$，对应的方程组为：$\left\{\begin{array}{l}{1•x+0•y=3}\\{0•x+1•y=1}\end{array}\right.$，则$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$，代入方程组，即可求得m和n的值，即可求得矩阵$（\begin{array}{l}{m}\\{n}\end{array}）$的值．

解答 解：矩阵为$（\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{1}\end{array}）$，对应的方程组为：$\left\{\begin{array}{l}{1•x+0•y=3}\\{0•x+1•y=1}\end{array}\right.$，解得：$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$，
由题意得：关于x、y的二元线性方程组$\left\{\begin{array}{l}{2x+my=5}\\{nx-4y=2}\end{array}\right.$的解为：$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$，
∴$\left\{\begin{array}{l}{2×3+m=5}\\{3n-4×1=2}\end{array}\right.$，解得：$\left\{\begin{array}{l}{m=-1}\\{n=2}\end{array}\right.$，
$（\begin{array}{l}{m}\\{n}\end{array}）$=$（\begin{array}{l}{-1}\\{2}\end{array}）$，
故答案为：$（\begin{array}{l}{-1}\\{2}\end{array}）$．

点评 本题的考点是二元一次方程组的矩阵形式，主要考查了几种特殊的矩阵变换，解答的关键是对增广矩阵的理解，利用方程组同解解决问题，属于基础题．