Question

20．关于x，y的方程组$\left\{{\begin{array}{l}{3ax+2y-1=0}\\{x+ay+3=0}\end{array}}\right.$的增广矩阵是$（\begin{array}{cc}3a&2\\ 1&a\end{array}\right.\begin{array}{c}1\\-3\end{array}）\right.$．

Answer 1

分析 先把方程组方程组$\left\{\begin{array}{l}3ax+2y-1=0\\ x+ay+3=0\end{array}\right.$改写为$\left\{\begin{array}{l}3ax+2y=1\\ x+ay=-3\end{array}\right.$，再由增广矩阵的概念进行求解．

解答 解：二元一次方程组$\left\{\begin{array}{l}3ax+2y-1=0\\ x+ay+3=0\end{array}\right.$，即$\left\{\begin{array}{l}3ax+2y=1\\ x+ay=-3\end{array}\right.$，
∴二元一次方程组$\left\{\begin{array}{l}3ax+2y=1\\ x+ay=-3\end{array}\right.$的增广矩阵是$（\begin{array}{cc}3a&2\\ 1&a\end{array}\right.\begin{array}{c}1\\-3\end{array}）\right.$
$（\begin{array}{cc}3a&2\\ 1&a\end{array}\right.\begin{array}{c}1\\-3\end{array}）\right.$，
故答案为：$（\begin{array}{cc}3a&2\\ 1&a\end{array}\right.\begin{array}{c}1\\-3\end{array}）\right.$

点评 本题考查二元一次方程组的矩阵形式，是基础题，解题时要认真审题，注意熟练掌握增广矩阵的概念．