Answer
If the matrix equation $AX=B$ has a unique solution, then we can solve the
Equation using $X={{A}^{-1}}B$.
Work Step by Step
The matrix $AX=B$ is determined by considering the simultaneous equations.
$\begin{align}
& x+2y=4 \\
& 3x-5y=1 \\
\end{align}$
Now, we can write in matrix form as below:
$\left[ \left. \begin{matrix}
1 & 2 \\
3 & -5 \\
\end{matrix} \right|\begin{matrix}
x \\
y \\
\end{matrix} \right]=\left[ \begin{align}
& 4 \\
& 1 \\
\end{align} \right]$
Where, $A=\left[ \begin{matrix}
1 & 2 \\
3 & -5 \\
\end{matrix} \right]$ $X=\left[ \begin{align}
& X \\
& Y \\
\end{align} \right]$ $B=\left[ \begin{align}
& 4 \\
& 1 \\
\end{align} \right]$
Now, we will solve the matrix $AX=B$.
Multiply $X$, $A$ and the inverse of $B$ which is not possible because there is a formula to find the $X$.
$AX=B$
So, multiply both sides by the inverse of $A$.
${{A}^{-1}}AX={{A}^{-1}}B$
And, we have ${{A}^{-1}}A=I$ which is the identity matrix. Therefore $IX=X$
So, $X={{A}^{-1}}B$
Thus, there is a formula for finding the value of $X$.
