I wrote the system of equations in the form AX = B, however how would I go about calculating the solution using the equation X = A^-1B?
Go to this web page.
Scroll about half way down.
You will find material of inverses of a $\displaystyle 2\times 2$ matrix.
You would find the inverse of the matrix and multiply by it! That is exactly what $\displaystyle A^{-1}B$ means.
The inverse of the two by two matrix [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix} is $\displaystyle \frac{1}{\Delta}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$ where $\displaystyle \Delta= ad- bc$, the determinant of the matrix. Be sure to do the multiplication in the order shown.
if $\displaystyle AX = B$
then $\displaystyle A^{-1}(AX) = A^{-1}B$ so
$\displaystyle (A^{-1}A)X = A^{-1}B$, that is:
$\displaystyle IX = X = A^{-1}B$.
so if A is indeed invertible, then calculating $\displaystyle A^{-1}B$ tells you which vector "X" is.