I realize that this may not be exactly what you are looking for, but I always found it a very nice simple example illustrating the usefulness of matrices.
Consider the following problem (with parameters that I made up): Every year people get the flu. Those who get the flu this year have a $30\%$ chance of getting the flu next year, whereas those who do not get the flu this year have a $40\%$ chance of getting the flu next year. This problem can be modelled by the matrix (why?)
$$
M=\left(\begin{matrix}0.3&0.4\\0.7&0.6\end{matrix}\right).
$$
Supposing that this year there were $x$ people who got the flu and $y$ people who did not get the flu, we can now give an estimate for the number of people, say $x_n$ who get the flu after $n$ years, and the number of people, say $y_n$, who do not get the flu, namely
$$
\left(\begin{matrix}x_n\\y_n\end{matrix}\right)=\left(\begin{matrix}0.3&0.4\\0.7&0.6\end{matrix}\right)^n\left(\begin{matrix}x\\y\end{matrix}\right).
$$
Now, matrix multiplication is a laborious process unless the matrices have a simple form, say a diagonal form for instance. Luckily, there is basis which diagonalizes this matrix, i.e. we can pick an invertible matrix $U$ such that
$$
\left(\begin{matrix}0.3&0.4\\0.7&0.6\end{matrix}\right)=U^{-1}\left(\begin{matrix}\lambda_1 &0\\0&\lambda_2\end{matrix}\right)U,
$$
and this is something you can show by methods relying on the determinant. It follows that
$$
\left(\begin{matrix}x_n\\y_n\end{matrix}\right)=U^{-1}\left(\begin{matrix}\lambda_1^n&0\\0&\lambda_2^n\end{matrix}\right)U\left(\begin{matrix}x\\y\end{matrix}\right).
$$
With this much simpler form, it is possible to answer questions such as whether or not the population 'stabilizes' in the sense that there are $x_\infty,y_\infty$ such that $x_n\rightarrow x_\infty,y_n\rightarrow y_\infty$, as one might expect.
