Answer
$(-2,1)$
Work Step by Step
This problem is connected to exercise $35$, as the coefficient matrix of the system
$ \left\{\begin{array}{l}{2 x+y=-3}\\{ax+ay=-a}\end{array}\right., a\neq 0$
is
$A=\left[\begin{array}{ll}{2}&{1}\\{a}&{a}\end{array}\right]$,
and we found
$A^{-1}=\displaystyle \left[\begin{array}{rr}{1}&{-\displaystyle \frac{1}{a}}\\{-1}&{\displaystyle \frac{2}{a}}\end{array}\right]$
Writing the system in matrix form, $AX=B, $ the solution is
$X=A^{-1}B$
$\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{rr}
{1}&{-\displaystyle \frac{1}{a}}\\
{-1}&{\displaystyle \frac{2}{a}}\end{array}\right]\left[\begin{array}{l}
-3\\
-a
\end{array}\right]$
$\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{l}
-3+1\\
3-2
\end{array}\right]=\left[\begin{array}{l}
-2\\
1
\end{array}\right]$
Solution: $(x,y)=(-2,1)$
