$解:把\begin{cases}{x=1}\\{y=-1}\end{cases}代入方程组$
$得\begin{cases}{a+b=2}\\{c+3=-2}\end{cases},所以\begin{cases}{a+b=2}\\{c=-5}\end{cases}$
$把\begin{cases}{x=2}\\{y=-6}\end{cases}代入ax-by=2$
$得2a+6b=2,即a+3b=1$
$所以\begin{cases}{a+b=2}\\{a+3b=1}\end{cases},解得\begin{cases}{a=\frac{5}{2}}\\{b=-\frac{1}{2}}\end{cases}$
$所以a=\frac{5}{2},b=-\frac{1}{2},c=-5$