How to use matrices to solve a circuit
The schematic, if I'm reading your problem correctly, is this:



simulate this circuit &ndash; Schematic created using CircuitLab
If I'm right then you happen to know, a priori, two of the node voltages. But let's say, for now, that we want to write out the nodal equations for all four nodes:
$$\begin{array}{ccccl}
V_1\cdot G_1&amp;-V_2\cdot G_1&amp;+V_3\cdot 0&amp;+V_4\cdot 0&amp;=I_1\\
-V_1\cdot G_1 &amp;+ V_2\cdot\left(G_1+G_2+G_3\right)&amp;-V_3\cdot G_3&amp;+V_4\cdot 0&amp;=0\:\text{A}\\
V_1\cdot 0 &amp;-V_2\cdot G_3 &amp;+ V_3\cdot\left(G_3+G_4+G_5\right)&amp;-V_4\cdot G_5 &amp;= 0\:\text{A}\\
V_1\cdot 0 &amp;+V_2\cdot 0&amp;-V_3\cdot G_5&amp;+V_4\cdot G_5&amp;=-I_2
\end{array}$$
This is the same as:
$$\left[\begin{smallmatrix}
G_1&amp;-G_1&amp;0&amp;0\\
-G_1 &amp;G_1+G_2+G_3&amp;-G_3&amp; 0\\
0 &amp;-G_3 &amp; G_3+G_4+G_5&amp;-G_5\\
0 &amp;0&amp;-G_5&amp;G_5
\end{smallmatrix}\right]\left[\begin{smallmatrix}V_1\\V_2\\V_3\\V_4\end{smallmatrix}\right]=\left[\begin{smallmatrix}I_1\\0\:\text{A}\\0\:\text{A}\\-I_2\end{smallmatrix}\right]$$
You already know \$V_1\$ and \$V_4\$, so this becomes:
$$\left[\begin{smallmatrix}
\frac12\:\text{S}&amp;-\frac12\:\text{S}&amp;0\:\text{S}&amp;0\:\text{S}\\
-\frac12\:\text{S} &amp;\frac{11}{12}\:\text{S}&amp;-\frac16\:\text{S}&amp; 0\:\text{S}\\
0\:\text{S} &amp;-\frac16\:\text{S} &amp; \frac35\:\text{S}&amp;-\frac1{10}\:\text{S}\\
0\:\text{S} &amp;0\:\text{S}&amp;-\frac1{10}\:\text{S}&amp;\frac1{10}\:\text{S}
\end{smallmatrix}\right]\left[\begin{smallmatrix}8\:\text{V}\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\-1\:\text{V}\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\0\:\text{A}\phantom{\frac{11}{12}}\\0\:\text{A}\phantom{\frac{11}{12}}\\-I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]$$
So, this is four equations in four unknowns. Do you know how to re-arrange it and then solve?
One step is to get all the unknowns in the same column-array. You can do this easily by pulling the unknowns to the left. So, verify that you can see how to re-arrange it this way:
$$\left[\begin{smallmatrix}
1&amp;\frac12\:\text{S}&amp;0\:\text{S}&amp;0\\
0 &amp;-\frac{11}{12}\:\text{S}&amp;\frac16\:\text{S}&amp; 0\\
0 &amp;\frac16\:\text{S} &amp; -\frac35\:\text{S}&amp;0\\
0 &amp;0\:\text{S}&amp;-\frac1{10}\:\text{S}&amp;1
\end{smallmatrix}\right]\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix}4\:\text{A}\phantom{\frac{11}{12}}\\-4\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\end{smallmatrix}\right]$$
Once you are here, you can use any or all of the usual methods of hand-solutions. These include Cramer's Rule or diagonalizing the left side matrix, etc. Lots of ways to go.
Of course, the canonical way to go is this:
$$\left[\begin{smallmatrix}
1&amp;\frac12\:\text{S}&amp;0\:\text{S}&amp;0\\
0 &amp;-\frac{11}{12}\:\text{S}&amp;\frac16\:\text{S}&amp; 0\\
0 &amp;\frac16\:\text{S} &amp; -\frac35\:\text{S}&amp;0\\
0 &amp;0\:\text{S}&amp;-\frac1{10}\:\text{S}&amp;1
\end{smallmatrix}\right]^{-1}\left[\begin{smallmatrix}
1&amp;\frac12\:\text{S}&amp;0\:\text{S}&amp;0\\
0 &amp;-\frac{11}{12}\:\text{S}&amp;\frac16\:\text{S}&amp; 0\\
0 &amp;\frac16\:\text{S} &amp; -\frac35\:\text{S}&amp;0\\
0 &amp;0\:\text{S}&amp;-\frac1{10}\:\text{S}&amp;1
\end{smallmatrix}\right]\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix}
1&amp;\frac12\:\text{S}&amp;0\:\text{S}&amp;0\\
0 &amp;-\frac{11}{12}\:\text{S}&amp;\frac16\:\text{S}&amp; 0\\
0 &amp;\frac16\:\text{S} &amp; -\frac35\:\text{S}&amp;0\\
0 &amp;0\:\text{S}&amp;-\frac1{10}\:\text{S}&amp;1
\end{smallmatrix}\right]^{-1}\left[\begin{smallmatrix}4\:\text{A}\phantom{\frac{11}{12}}\\-4\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\end{smallmatrix}\right]\\\\\therefore\\\\
\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix}
1&amp;\frac12\:\text{S}&amp;0\:\text{S}&amp;0\\
0 &amp;-\frac{11}{12}\:\text{S}&amp;\frac16\:\text{S}&amp; 0\\
0 &amp;\frac16\:\text{S} &amp; -\frac35\:\text{S}&amp;0\\
0 &amp;0\:\text{S}&amp;-\frac1{10}\:\text{S}&amp;1
\end{smallmatrix}\right]^{-1}\left[\begin{smallmatrix}4\:\text{A}\phantom{\frac{11}{12}}\\-4\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\end{smallmatrix}\right]
$$

It's been two days, now. So I may as well provide the detailed solution for others to consider:
$$\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix}
1&amp;\frac12\:\text{S}&amp;0\:\text{S}&amp;0\\
0 &amp;-\frac{11}{12}\:\text{S}&amp;\frac16\:\text{S}&amp; 0\\
0 &amp;\frac16\:\text{S} &amp; -\frac35\:\text{S}&amp;0\\
0 &amp;0\:\text{S}&amp;-\frac1{10}\:\text{S}&amp;1
\end{smallmatrix}\right]^{-1}\left[\begin{smallmatrix}4\:\text{A}\phantom{\frac{11}{12}}\\-4\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\end{smallmatrix}\right]\\\\\therefore\\\\
\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\\\V_2\phantom{\frac{11}{12}}\\\\V_3\phantom{\frac{11}{12}}\\\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix}\frac{323}{188}\\\\ \frac{429}{94}\\\\ \frac{207}{188}\\\\ \frac{79}{376}\end{smallmatrix}\right]\approx \left[\begin{smallmatrix}1.71808510638298\phantom{\frac{11}{12}}\\\\ 4.56382978723404\phantom{\frac{11}{12}}\\\\ 1.10106382978723\phantom{\frac{11}{12}}\\\\ 0.210106382978723\phantom{\frac{11}{12}}\end{smallmatrix}\right]
$$