I liked examples about finite state games.
For example imagine you have a game where there are three lights. These lights can only be on or off; and you have 3 buttons which will do different things to the lights.
We represent a light being on with the number $1$ and off with $0$.
We can then represent the action of pushing the buttons onto our configuration of lights with vectors inside $\mathbb{Z}_2^3$, that is vectors with $3$ components which have entries respecting arithmetic modulo $2$. You can see how this works in the following way. Imagine we have $3$ lights and they are side by side. The first light is on, the second is off, and the third is on. We can write this down like so:
$$\text{Light starting config}:=\mathbf{s}=\begin{pmatrix}1\\0\\1\end{pmatrix}.$$
Suppose further that we have access to a button that will switch the first and middle lights (by switch I mean if that light is on after pressing it the light will be off), and does nothing to the last light. This can be represented as:
$$\text{button}:=\mathbf{b}=\begin{pmatrix}1\\1\\0\end{pmatrix}.$$
To see why this is true if we add $\mathbf{b}$ to our light configuration what do we get:
$$\mathbf{b}+\mathbf{s}=\begin{pmatrix}1\\1\\0\end{pmatrix}+\begin{pmatrix}1\\0\\1\end{pmatrix}=\begin{pmatrix}0\\1\\1\end{pmatrix}=\text{Config after pressing}.$$
As would be expected our representation of pushing the button has changed the configuration of lights so that now the first light is off, the second is on, and the third is on.
What if we have more than one button? Can we make the lights reach a certain configuration with some specific set of buttons and a starting configuration?
We can use the preceding discussion and some more linear algebra to answer these questions. An example will show you what I mean.
So suppose we start with $3$ lights in the same position as before:
$$\mathbf{s}=\begin{pmatrix}1\\0\\1\end{pmatrix}.$$
And three buttons:
$$\mathbf{b}_1=\begin{pmatrix}1\\0\\0\end{pmatrix},\mathbf{b}_2=\begin{pmatrix}1\\1\\0\end{pmatrix},\mathbf{b}_3=\begin{pmatrix}1\\0\\1\end{pmatrix}$$
And we want to know is it possible that after pushing some sequence of buttons we can arrange the lights to be in the configuration: $$\mathbf{d}=\begin{pmatrix}0\\1\\0\end{pmatrix}.$$
If you think about it it doesn't matter which order we push the buttons in and so it matters only if we push the button at all. Therefore what we are asking in terms of our model is whether the following is true:
$$\text{Does there exist $c_1,c_2$ and $c_3$ in $\mathbb{Z}_2$ such that:}\quad c_1\mathbf{b}_1+c_2\mathbf{b}_2+c_3\mathbf{b}_3+\mathbf{s}=\mathbf{d}.$$
Here $c_1,c_2$ and $c_3$ represent if we push buttons $1,2$ and $3$ respectively. We can rewrite the equation as:
$$c_1\mathbf{b}_1+c_2\mathbf{b}_2+c_3\mathbf{b}_3=\mathbf{d}-\mathbf{s}$$
$$\downarrow$$
$$\begin{pmatrix}c_1 & c_2 & c_3\\0 & c_2 & 0\\0 & 0 & c_3\end{pmatrix}=\begin{pmatrix}1\\1\\1\end{pmatrix}$$
After solving this in the usual way (except remember now we are doing arithmetic modulo $2$) you can see if we push each button once we will have put the lights into their desired configuration. If this matrix had no solution then we would not be able to put them into the given configuration.