Answer
$(-5,10)$
Work Step by Step
Solving $AX=B,\ \qquad X=A^{-1}B$.
This problem is connected to exercise $31$, as the coefficient matrix is
$A=\left[\begin{array}{ll}{2}&{1}\\{1}&{1}\end{array}\right]$, and $A^{-1}=\left[\begin{array}{rr}{1}&{-1}\\{-1}&{2}\end{array}\right]$
Writing the system $\left\{\begin{array}{r}{2 x+y=0}\\{x+y=5}\end{array}\right.$
in matrix form, $AX=B, $the solution is
$X=A^{-1}B=\left[\begin{array}{rr}
{1}&{-1}\\
{-1}&{2}\end{array}\right]\left[\begin{array}{l}
0\\
5
\end{array}\right]$
$\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{l}
0-5\\
0+10
\end{array}\right]=\left[\begin{array}{l}
-5\\
10
\end{array}\right]$
Solution: $(x,y)=(-5,10)$
