I'm quite confused on using this matrix equation to find the voltage across the 3ohm resistor. Once I fill in the values of the resistors for the Y values where Y=1/Z I get stuck. I'm not quite sure how to continue with it any help or tips would be greatly appreciated. I have to use the given matrix equations in the form YV = I
Best Answer
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:
$$\left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\V_2\phantom{\frac{11}{12}}\\V_3\phantom{\frac{11}{12}}\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix} 1&\frac12\:\text{S}&0\:\text{S}&0\\ 0 &-\frac{11}{12}\:\text{S}&\frac16\:\text{S}& 0\\ 0 &\frac16\:\text{S} & -\frac35\:\text{S}&0\\ 0 &0\:\text{S}&-\frac1{10}\:\text{S}&1 \end{smallmatrix}\right]^{-1}\left[\begin{smallmatrix}4\:\text{A}\phantom{\frac{11}{12}}\\-4\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\\\frac1{10}\:\text{A}\phantom{\frac{11}{12}}\end{smallmatrix}\right]\\\\\therefore\\\\ \left[\begin{smallmatrix}I_1\phantom{\frac{11}{12}}\\\\V_2\phantom{\frac{11}{12}}\\\\V_3\phantom{\frac{11}{12}}\\\\I_2\phantom{\frac{11}{12}}\end{smallmatrix}\right]=\left[\begin{smallmatrix}\frac{323}{188}\\\\ \frac{429}{94}\\\\ \frac{207}{188}\\\\ \frac{79}{376}\end{smallmatrix}\right]\approx \left[\begin{smallmatrix}1.71808510638298\phantom{\frac{11}{12}}\\\\ 4.56382978723404\phantom{\frac{11}{12}}\\\\ 1.10106382978723\phantom{\frac{11}{12}}\\\\ 0.210106382978723\phantom{\frac{11}{12}}\end{smallmatrix}\right] $$