There is an elementary problem that goes more or less like this: you have a special telephone keyboard with nine lighted buttons (one for each number from $1$ to $9$); when pushing each button other than number $5$ (the central button) then this switches the state of the lights of the button itself and of all its surrounding buttons; pushing number $5$ only switches the state of the lights of its surrounding buttons, but not of itself. Starting with all lights off, the question asks whether we can get all lights on by pushing buttons. The obvious solution to the negative answer relies on the fact that the parity of lighted buttons at every state of the keyboard is an invariant. But there is also a sophisticated solution.
Take the set $X$ of $9$ elements and think of $\mathcal{P}(X)$ as a vector space over the field $\mathbb{Z}_2$ with the sum being the symmetric difference and the product given by $0.v=\emptyset$ and $1.v=v$. Then we can identify each state of the keyboard with a corresponding vector in this space, while pushing the button $i$ corresponds to summing a special vector $v_i$ (associated to the button) to the vector representing the state of the keyboard. Thus, we are wondering if there are some scalars $\alpha_i$ such that $\sum_{i=1}^{9} \alpha_iv_i=X$. Writing each $v_i$ and $X$ in the base of the space given by the singleton elements $1, ..., 9$, we get a system of linear equations which can be seen to have no solutions by computing the $9 \times 9$ determinant and verifying it is null.