I search in internet and I only found square matrix examples.
It's possible use Cramer rule to resolve this matrix .
From what I see, I would have to use a different method to solve it, because I can not add the two columns next to the matrix, as the cramer rule says.
Independent terms $$ \begin{bmatrix} 11 \\ 9 \\ 53 \\ 15\\ \end{bmatrix} $$
Matrix $$ \begin{bmatrix} 3 & -1 & 2 \\ -2 & 3 & 8 \\ -7 & 5 & -6 \\ 4 & -1 & 0\\ \end{bmatrix} $$