Answer
$a.\qquad\left[\begin{array}{ll}
5 & 3\\
3 & 2
\end{array}\right]\cdot\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{l}
4\\
3
\end{array}\right]$
$b.\qquad A^{-1}=\left[\begin{array}{ll}
2 & -3\\
-3 & 5
\end{array}\right]$
$c.\qquad\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{ll}
2 & -3\\
-3 & 5
\end{array}\right]\left[\begin{array}{l}
4\\
3
\end{array}\right]=\left[\begin{array}{l}
-1\\
3
\end{array}\right]$
$d.\qquad x=-1,\qquad y=3$
Work Step by Step
$a.$
$A\cdot X=B$
$\left[\begin{array}{ll}
5 & 3\\
3 & 2
\end{array}\right]\cdot\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{l}
4\\
3
\end{array}\right]$
$(b)$
$A=\displaystyle \left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right] \Rightarrow A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{ll}
d & -b\\
-c & a
\end{array}\right]$
$A^{-1}=\displaystyle \frac{1}{5(2)-3(3)}\left[\begin{array}{ll}
2 & -3\\
-3 & 5
\end{array}\right]$
$A^{-1}=\displaystyle \frac{1}{1}\left[\begin{array}{ll}
2 & -3\\
-3 & 5
\end{array}\right]$
$A^{-1}=\left[\begin{array}{ll}
2 & -3\\
-3 & 5
\end{array}\right]$
$(c)$
$X=A^{-1}B=\left[\begin{array}{ll}
2 & -3\\
-3 & 5
\end{array}\right]\left[\begin{array}{l}
4\\
3
\end{array}\right]=\left[\begin{array}{l}
2(4)-3(3)\\
-3(4)+5(3)
\end{array}\right]=\left[\begin{array}{l}
-1\\
3
\end{array}\right]$
$\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{ll}
2 & -3\\
-3 & 5
\end{array}\right]\left[\begin{array}{l}
4\\
3
\end{array}\right]=\left[\begin{array}{l}
-1\\
3
\end{array}\right]$
$(d)$
$x=-1,\qquad y=3$
