This might be better suited for the Physics Stack Exchange, if that's the case feel free to migrate it there. I'm also not terribly familiar with the tagging here on the Mathematics SE so if I tagged incorrectly, I'd appreciate any improvements.
I was studying for an upcoming undergraduate Physics exam when I found this slidedeck from another university. I don't attend Rochester, so I obviously wasn't at that lecture, but there's an interesting bit on Page 5 that notes that the system of equations used to solve the problem can be converted into a matrices to be solved using linear algebra. Now while my physics skills are pretty mediocre, I really enjoy linear algebra, so I was curious as to what the "fancy linear-algebra methods" mentioned on Slide 7 might be.
The slides say that it would involve finding the inverse of the 6x6 matrix, which I assume would entail putting it in an augmented matrix with the 6x6 Identity Matrix and row reducing, and then left multiplying both sides. That doesn't sound too fancy to me, although it may have simply been sarcasm on the professor's part.
Is there something I'm missing or is it just row reduction and left-multiplying?
Per Sean Roberson’s comment, the reference to “fancy linear-algebra methods” in the slides is likely a facetious reference to Gaussian elimination (a.k.a. row-reduction) or Cramer’s rule. There are, however, other fancy linear-algebra ways to solve such networks problems that likely weren’t what the lecturer had in mind.
If we view the resistive network as a two-dimensional complex, a current distribution in the wires is a function from branches to real numbers, so using the language of algebraic topology we can identify them with the space of 1-chains $C_1$. In a similar way, net current flows at nodes are 0-chains, voltage drops across branches are 1-cochains, and potentials at the nodes are 0-cochains. The boundary operator $\partial:C_1\to C_0$ acts as a bulk adder, that is, for any current distribution $\mathbf I$, $\partial\mathbf I$ is the corresponding net current flow at the nodes. In this framework, Kirchhoff’s current law is simply $\partial\mathbf I=0$. The coboundary operator $\mathrm{d}:C^0\to C^1$ acts as a bulk difference operator: given an assignment of potentials to nodes, it computes the corresponding voltage drops. In this framework, the most useful form of Kirchhoff’s voltage law is $\mathbf V=-\mathrm{d}\mathbf\Phi$, which basically says that one can make a consistent assignment $\mathbf\Phi$ of potentials to nodes, in the sense that if $\partial\alpha = B-A$, then $V^\alpha=\Phi^A-\Phi^B$. Finally, Ohm’s law connects $C_1$ and $C^1$: $\mathbf V=Z\mathbf I$. We can put these together into an equation that connects node potentials $\mathbf\Phi$ to net node currents $\mathbf K$: $$-\partial Z^{-1}\mathrm{d}\mathbf\Phi = \mathbf K.\tag{*}$$ The operator $-\partial Z^{-1}\mathrm{d}$ is the discrete Laplacian of the network that you might be familiar with from graph theory.
There are analogs of the mesh current and node potential methods in this framework, but the example resistive network is best treated as a Dirichlet problem: we’re given the values of $\mathbf\Phi$ at the boundary of the network and require that $\mathbf K=0$ on interior nodes. Specifically, labeling the nodes clockwise from the top, our initial conditions are $\Phi^A=V$, $\Phi^C=0$, $K_B=K_D=0$. Using the orientations in the diagram and expanding equation (*) in matrix form, we have $$\begin{bmatrix}-1&-1&0&0&0\\0&1&1&0&-1\\0&0&0&1&1\\1&0&-1&-1&0\end{bmatrix} \begin{bmatrix}\frac1{R_1}&0&0&0&0\\0&\frac1{R_2}&0&0&0\\0&0&\frac1{R_3}&0&0\\0&0&0&\frac1{R_4}&0\\0&0&0&0&\frac1{R_5}\end{bmatrix} \begin{bmatrix}-1&0&0&1\\-1&1&0&0\\0&1&0&-1\\0&0&1&-1\\0&-1&1&0\end{bmatrix} \begin{bmatrix}V\\\Phi^B\\0\\\Phi^D\end{bmatrix} = \begin{bmatrix}I\\0\\-I\\0\end{bmatrix}.$$ This matrix product looks messy, but it’s quite easy to compute piecemeal: the outer matrices produce simple sums of various elements of the vector, while the diagonal matrix in the middle scales each element. The net result is a system of four linear equations in three unknowns, but we can remove a redundant equation a priori by working with the reduced operators $[\partial]$ and $[\mathrm{d}]$: delete rows and columns that correspond to nodes at ground, in this case the third node $C$. You end up with the system $$\begin{align} -\frac 1{R_2}\Phi^B-\frac 1{R_1}\Phi^D + \frac V{R_1}+\frac V{R_2} &= I \\ \left(\frac1{R_2}+\frac1{R_3}+\frac1{R_5}\right)\Phi^B - \frac1{R_3}\Phi^D-\frac V{R_2} &= 0 \\ -\frac1{R_3}\Phi^B +\left(\frac1{R_1}+\frac1{R_3}+\frac1{R_4}\right)\Phi^D-\frac V{R_1} &= 0 \end{align}$$ which I think looks a little less daunting than the one in the slides, especially if you replace resistance with conductance ($G_\alpha=\frac1{R_\alpha}$). Solving this system produces the same value for $I$ as was demonstrated in the slides, and if you want to recover the individual voltage drops or currents in the wires, they are $-\mathrm{d}\mathbf\Phi$ and $-Z^{-1}\mathrm{d}\mathbf\Phi$, respectively.
I generally like this method more than the ad-hoc one demonstrated in the slides because you proceed pretty much mechanically from a description of the network—the $\partial$ and $\mathrm d$ operators together with resistnces $Z$—to a relatively small set of equations to be solved. You don’t have to pick out loops and hope that you’ve selected enough independent ones. The machinery of this method basically does that for you automatically. If you allow complex values, then you can accommodate inductors and capacitors in this framework, but you’re no longer guaranteed a solution as you are with pure linear resistive networks. Those no-solution cases generally correspond to resonances.