How to use matrices to solve a circuit

The schematic, if I'm reading your problem correctly, is this:

^{simulate this circuit – 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:

$$\begin{array}{ccccl} V_1\cdot G_1&-V_2\cdot G_1&+V_3\cdot 0&+V_4\cdot 0&=I_1\\ -V_1\cdot G_1 &+ V_2\cdot\left(G_1+G_2+G_3\right)&-V_3\cdot G_3&+V_4\cdot 0&=0\:\text{A}\\ V_1\cdot 0 &-V_2\cdot G_3 &+ V_3\cdot\left(G_3+G_4+G_5\right)&-V_4\cdot G_5 &= 0\:\text{A}\\ V_1\cdot 0 &+V_2\cdot 0&-V_3\cdot G_5&+V_4\cdot G_5&=-I_2 \end{array}$$

This is the same as:

$$\left[\begin{smallmatrix} G_1&-G_1&0&0\\ -G_1 &G_1+G_2+G_3&-G_3& 0\\ 0 &-G_3 & G_3+G_4+G_5&-G_5\\ 0 &0&-G_5&G_5 \end{smallmatrix}\right]\left[\begin{smallmatrix}V_1\\V_2\\V_3\\V_4\end{smallmatrix}\right]=\left[\begin{smallmatrix}I_1\\0\:\text{A}\\0\:\text{A}\\-I_2\end{smallmatrix}\right]$$

You already know \$V_1\$ and \$V_4\$, so this becomes:

$$\left[\begin{smallmatrix} \frac12\:\text{S}&-\frac12\:\text{S}&0\:\text{S}&0\:\text{S}\\ -\frac12\:\text{S} &\frac{11}{12}\:\text{S}&-\frac16\:\text{S}& 0\:\text{S}\\ 0\:\text{S} &-\frac16\:\text{S} & \frac35\:\text{S}&-\frac1{10}\:\text{S}\\ 0\:\text{S} &0\:\text{S}&-\frac1{10}\:\text{S}&\frac1{10}\:\text{S} \end{smallmatrix}\right]\left[\begin{smallmatrix}8\:\text{V}\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\-1\:\text{V}\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\0\:\text{A}\phantom{\frac{11}{12}}\\0\:\text{A}\phantom{\frac{11}{12}}\\-I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]$$

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:

$$\left[\begin{smallmatrix} 1&\frac12\:\text{S}&0\:\text{S}&0\\ 0 &-\frac{11}{12}\:\text{S}&\frac16\:\text{S}& 0\\ 0 &\frac16\:\text{S} & -\frac35\:\text{S}&0\\ 0 &0\:\text{S}&-\frac1{10}\:\text{S}&1 \end{smallmatrix}\right]\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix}4\:\text{A}\phantom{\frac{11}{12}}\\-4\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\end{smallmatrix}\right]$$

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:

$$\left[\begin{smallmatrix} 1&\frac12\:\text{S}&0\:\text{S}&0\\ 0 &-\frac{11}{12}\:\text{S}&\frac16\:\text{S}& 0\\ 0 &\frac16\:\text{S} & -\frac35\:\text{S}&0\\ 0 &0\:\text{S}&-\frac1{10}\:\text{S}&1 \end{smallmatrix}\right]^{-1}\left[\begin{smallmatrix} 1&\frac12\:\text{S}&0\:\text{S}&0\\ 0 &-\frac{11}{12}\:\text{S}&\frac16\:\text{S}& 0\\ 0 &\frac16\:\text{S} & -\frac35\:\text{S}&0\\ 0 &0\:\text{S}&-\frac1{10}\:\text{S}&1 \end{smallmatrix}\right]\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix} 1&\frac12\:\text{S}&0\:\text{S}&0\\ 0 &-\frac{11}{12}\:\text{S}&\frac16\:\text{S}& 0\\ 0 &\frac16\:\text{S} & -\frac35\:\text{S}&0\\ 0 &0\:\text{S}&-\frac1{10}\:\text{S}&1 \end{smallmatrix}\right]^{-1}\left[\begin{smallmatrix}4\:\text{A}\phantom{\frac{11}{12}}\\-4\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\end{smallmatrix}\right]\\\\\therefore\\\\ \left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix} 1&\frac12\:\text{S}&0\:\text{S}&0\\ 0 &-\frac{11}{12}\:\text{S}&\frac16\:\text{S}& 0\\ 0 &\frac16\:\text{S} & -\frac35\:\text{S}&0\\ 0 &0\:\text{S}&-\frac1{10}\:\text{S}&1 \end{smallmatrix}\right]^{-1}\left[\begin{smallmatrix}4\:\text{A}\phantom{\frac{11}{12}}\\-4\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\end{smallmatrix}\right] $$

It's been two days, now. So I may as well provide the detailed solution for others to consider:

$$\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix} 1&\frac12\:\text{S}&0\:\text{S}&0\\ 0 &-\frac{11}{12}\:\text{S}&\frac16\:\text{S}& 0\\ 0 &\frac16\:\text{S} & -\frac35\:\text{S}&0\\ 0 &0\:\text{S}&-\frac1{10}\:\text{S}&1 \end{smallmatrix}\right]^{-1}\left[\begin{smallmatrix}4\:\text{A}\phantom{\frac{11}{12}}\\-4\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\end{smallmatrix}\right]\\\\\therefore\\\\ \left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\\\V_2\phantom{\frac{11}{12}}\\\\V_3\phantom{\frac{11}{12}}\\\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix}\frac{323}{188}\\\\ \frac{429}{94}\\\\ \frac{207}{188}\\\\ \frac{79}{376}\end{smallmatrix}\right]\approx \left[\begin{smallmatrix}1.71808510638298\phantom{\frac{11}{12}}\\\\ 4.56382978723404\phantom{\frac{11}{12}}\\\\ 1.10106382978723\phantom{\frac{11}{12}}\\\\ 0.210106382978723\phantom{\frac{11}{12}}\end{smallmatrix}\right] $$

Forget about the matrices for now.

Write our your mesh equations for each loop. Each equation will have one term in it representing a net voltage source within the loop which may or may not be zero. All other terms will be IZ.

Now you want to solve these equations simultaneously and matrices are just a mindless way to do that.

So you first have to makes all the equations of the same form as each other so they fit into the matrix. In other words, make all the equations have the same terms by adding on terms to each equation that are present in other equations set I to zero for the terms that do not apply. That makes all the equations have all the same terms even if they are not applicable.

Then put those coefficients into the matrix and solve it. One column represents the net voltage, all other columnns represent the current through a particular Z. Then you row-reduce the matrix to solve it. The matrix is just a convenient way to solve simultaneous equations in this case.

Y*V=I . Here you know Y and I. Often what you would like to know is V, which is all the voltages at the nodes of the circuit. So, you solve this equation for V. To solve a matrix equation like this, it is easiest to use a computer program (like Matlab).

Kelko: Oh, my bad!!! I did not pay attention to this problem. It is really a trick question, since the equations given are not really applicable to this circuit. Sorry.

In this circuit, ask yourself what you know and what you want to know. There are four nodes in the circuit, but there are not four unknowns in this problem! In fact, you already know one node voltage is 8v and one node voltage is -1v. So, you will need to find two equations that will allow you to solve for v2 and v3, the two unknowns. No use writing kircoffs current law at the nodes with the voltage sources, because there is no way you can determine the currents that these sources may be generating. The two equations that you need are kircoffs current law at the the v2 and v3 nodes. So, this is not a problem with 4 equations in 4 unknowns, it is a problem with two equations in two unknowns. ( easy to solve without a computer)

If \$YV = I\$ is your given system, and you are looking for the currents through the resistors (=vector \$I\$) you simply have to multiply matrix \$Y\$ by vector \$V\$ (given voltages).

When you know the current through the resistor, you get the voltage across by Ohm's law.

The matrix equation is what most of the circuit calculation software use to solve circuits. It's derived from the KCL with application of Ohm's law to substitute \$I_{ab} = 1/Z_{ab}(V_a-V_b)\$ for all branches with impedance. One node is usually selected as "ground" where voltage is zero and so you have no equation for it, but have admittance to ground in the diagonal elements of the matrix. If a circuit has current sources they are accounted in the KCL. If a circuit has a voltage source, it's not an issue because the voltage source produce some current, so we just treat the current as unknown and write the KCL anyway.

Back to your problem.

You may notice that in your equations the 2nd and the 3rd may be solved separately to find \$V_2\$ and \$V_3\$. System of two equations is easy to solve by hand. Then you may as well solve the rest of the equations to find \$I_1\$ and \$I_2\$.

What happens is you effectively replace your left and right source with their Norton's equivalents to get a simple circuit with two unknown voltages.