Your matrix is incorrect for the formulated equations. The simplified equations are:
\$-7V_1+4V_2+3V_3=132\$
\$V_1-V_2-3V_2-3i_x=-9\$
\$5V_1-9V_3+20i_x=-500\$
\$V_3-V_2=22\$
Putting these equations into matrix form yields:
$$ \begin{bmatrix} -7 & 4 & 3 & 0\ \Omega \\ 1 & -1 & -3 & -3\ \Omega \\ 5 & 0 & -9 & 20\ \Omega \\ 0 & -1 & 1 & 0\ \Omega \end{bmatrix} \cdot \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ i_x \end{pmatrix} = \begin{pmatrix} 132 \text{ V} \\ -9 \text{ V} \\ -500 \text{ V} \\ 22 \text{ V} \end{pmatrix} $$
Solving this in Matlab gives the following X matrix (for AX=B):
X = $$\begin{bmatrix} -17.2619\\ -7.8333\\ 14.1667\\ -14.3095\end{bmatrix}$$
To verify, let's subtract \$V_2\$ from \$V_3\$. Here, it is: 14.1667 - (-7.8333) which is 22 V! This validates our solution. You can try this for the other voltages as well and see if the equations are satisfied.