题目内容

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$(\begin{array}{cc}3a&2\\ 1&a\end{array}\right.\begin{array}{c}1\\-3\end{array})\right.$.

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.

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