Answer
$$A=\left[\begin{array}{cc}{2} &{-1} \\{3}& {-2} \end{array}\right].$$
Work Step by Step
Assume that the matrix $A$ is given by
$$A=\left[\begin{array}{cc}{x_1} &{x_2} \\{x_3}& {x_4} \end{array}\right],$$
then we have the system
$$
\begin{aligned}
2x_{1}-x_{3}&=1 \\
2x_{2} -x_4 &=0\\
3x_1-2x_{3} &=0\\
3x_{2} -2x_4 &=1
\end{aligned}.
$$
The augmented matrix is given by
$$
\left[ \begin {array}{ccccc} 2&0&-1&0&1\\ 0&2&0&-1&0
\\ &0&-2&0&0\\ 0&3&0&-2&1
\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[ \begin {array}{ccccc} 1&0&0&0&2\\ 0&1&0&0&-1
\\ 0&0&1&0&3\\ 0&0&0&1&-2
\end {array} \right]
.$$
From which the solution is
$$x_1=2, \quad x_2=-1, \quad x_3=3, \quad x_4=-2 .$$
Hence, we have
$$A=\left[\begin{array}{cc}{2} &{-1} \\{3}& {-2} \end{array}\right].$$
