I have two good examples, they are somewhat similar, but one is solved more analytically, and the other more computationally:
Masses on a spring:
From classical mechanics: a mass on a spring with one end fixed can be described by the classical equation:
$ \mathbf{F} = m\ddot{\mathbf{x}} = -k\mathbf{x} $
If we have two masses on a spring, (in 1d)
$ m_1 \ddot{x}_1 = -k(|x_1 -x_2| - \ell_{\text{eq}})$
$ m_2 \ddot{x}_2 = -k(|x_1 -x_2| - \ell_{\text{eq}})$
When you get to more springs, this is more easily posed as a linear algebra problem:
$ \mathbf{M} \ddot{\mathbf{x}} = \mathbf{K}\mathbf{x}$
Where now, $\mathbf{M}$, and $\mathbf{K}$ are matrices describing the masses of the different objects and the spring constants between them respectively, and $\mathbf{x}$ describes the positions of the various objects (with the equilibrium distance $\ell$ subsumed into it).
If you look at this as a simple eigenvalue equation:
$ \ddot{\mathbf{x}} = \mathbf{M}^{-1}\mathbf{K} \mathbf{x} $
The eigenvectors (remember that $x$ is really time dependent), describe the normal modes of the system, and the eigenvalues are the frequencies of these modes. Demonstrating this with a system of 3 or so masses isn't THAT hard, and is very cool.
Schrödinger Equation steady state:
This equation is so fundamental to physics that it can really drive home how ridiculously important linear algebra is to Physicists. Unfortunately, describing what the wave function represents can be difficult. The best way is to describe the square of the wave function as a probability distribution. Thankfully as well, the steady state problem is real, and you don't have to deal with complex numbers. We can write down the Schrödinger equation as follows:
$ {\cal H}\Psi(t) = \left(-\frac{\hbar^2}{2m} \boldsymbol{\nabla} + V(\mathbf{x}) \right) \Psi(t) = -i \hbar \partial_t \Psi(t)$
From here, we can choose a basis by separating out the time dependence:
$ \left(-\frac{\hbar^2}{2m} \boldsymbol{\nabla} + V(\mathbf{x}) \right) \Psi_e = e \Psi_e $
And then we can write:
$ {\cal H}\Psi_e = e \Psi_e $
Where ${\cal H}$ is the finite difference expression of $\Psi$ on a grid. For example, for the 1D problem with a simple harmonic oscillator:
$ {\cal H}\Psi = -\frac{\hbar^2}{2m h^2} \left(\Psi(x_{i-1}) - 2\Psi(x_{i} + \Psi(x_{i+1}) \right) + \frac{k}{2} x_i^2 \Psi(x_{i}) $
Where $i$ is an index on the evenly spaced discretized grid.
Then you can find the eigenvalues and eigenvectors (unfortunately, for anything on a reasonable grid, this requires too many points to do by hand). These eigenvalues describe the steady state energies of the system! And the eigenvectors describe the steady state wave functions! Interestingly enough, this particular problem describes vibrational levels in molecules remarkably well (especially for the lowest energy states).
