It can be verified that
A = \left [ \begin{matrix} 3 & -7 & -2 & 2 \\ -3 & 5 & 1 & 0 \\ 6 & -4 & 0 & -5 \\ -9 & 5 & -5 & 12 \end{matrix} \right ] = \left [ \begin{matrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 2 & -5 & 1 & 0 \\ -3 & 8 & 3 & 1 \end{matrix} \right ] \left [ \begin{matrix} 3 & -7 & -2 & 2 \\ 0 & -2 & -1 & 2 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \end{matrix} \right ] = LU
Use this LU factorization of A to solve Ax = b, where b = \left [ \begin{matrix} -9 \\ 5 \\ 7 \\ 11 \end{matrix} \right ] .