Is this another way of saying the Transfer function is the output/input wherein my output is across [3,1] and input is across [1,1] ? I didn't get this part: "V1, Vx and Va are the first column of the inverse".
Remember how I ended up with the equations in this form:
If the 3 nodes along the top of the circuit, left to right, are labeled V1, Vx, Va, you can write the 3 node equations with a TestCurrent of 1 amp applied to the input:
(V1-Vx)*(s C1) = TestCurrent
(Vx - V1)*(s C1) + Vx*(1/(s L1)) + (Vx - Va)*(s C2) = 0
(Va-Vx)*(s C2) + Va*(1/R1) = 0
Collect terms and you will have:
V1*(s C1) + Vx*(-s C1) + Va*(0) = 1 The 1 here is the test current.
V1*(-s C1) + Vx*(s(C1+C2)+1/(s L1)) + Va*(-s C2) = 0
V1*(0) + Vx*(-s C2) + Va*(s C2 + 1/R1) = 0
Solve these equations for V1, Vx and Va using whatever method you like. I inverted the matrix of coefficients and the voltages V1, Vx and Va are the first column of the inverse; columns 2 and 3 are not needed. Form the expression Va/V1. Substitute numerical values and the rest follows.
They can be put into matrix form and then they look like this:
These equations could be solved by a linear equation solver, or by using Cramer's rule, or by brute force substitution and elimination; it's the student's choice. The result is then a column vector of the voltages at the three nodes when a current of 1 amp is injected at node V1:
However you could just as well inject the current at node 2, and solving would give the voltages at node V1, Vx and Va. The setup would look like this:
If you were to do that you would get some voltages that are irrelevant to the problem at hand, but it could be done.
Now, with this next setup you are solving for all the voltages at the three nodes, with a 1 amp current injected into node V1, into node Vx and into node Va, all in one setup. The result is a 3x3 matrix; it's the same result you get if you just invert the admittance matrix:
The first column is the result of the injection of 1 amp into V1. The second column gives the result of the 1 amp applied to Vx, and the 3rd column gives the result of 1 amp into node Va. The second and third columns are superfluous to the problem at hand, and we ignore them. It's that first column that you want.
Now the (1,1) element of the inverse matrix is the voltage at V1 when 1 amp is injected at V1, and the (3,1) element of the inverse is the voltage at Va. The inverse is a Z matrix and the ratio Va/V1, the desired transfer function, is given by Z(3,1)/Z(1,1).
Some may argue that taking the inverse and getting two extra unneeded columns is overkill. This is the 21st century and computing power is everywhere. Even my HP50G calculator can invert the matrix in a second, using only a few milliwatt-seconds of energy. Formulating the problem as a matrix and taking the inverse to get the desired Z(1,1) and Z(3,1) elements allows a very compact, quick and elegant solution to the problem.