I was watching my math lecture for a course and we were doing inverse matrices in finite fields in $\mathbb{Z}_5$. The determinant was $1 / (6-4)$ so $1/2$ which apparently means it is 3. I have no idea where 3 comes from, is $1$ over $2$ the same as division or does that mean something else?
I am entirely new to fields and have very limited knowledge on them, but my searches never revealed anything about division or what this would be for the determinant.
On
Division is defined the same way in modular arithmetic as it is on the regular number line. In other words, $\frac12$ is the unique number such that $2\cdot \frac12 = 1$. However, since modular arithmetic works differently from regular arithmetic, the interpretation of the above definition is very different: We have $2\cdot 3 = 1$, which means that $3$ actually fulfills the defining property of $\frac12$. We can also easily check that there are no other such numbers, so it is unique. That is why $\frac12 = 3$.
Note for later: If you ever have to solve quadratic equations in modular arithmetic (at least as long as it's modulo an odd prime), the formula $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ still works, except once again, the definition of $\sqrt{{}\cdot{}}$ is the same (sortof; there isn't anything like positive / negative to distinguish the two square roots of a number), but the interpretation is different, with, for instance, $\sqrt{-1} = \pm 2$ modulo $5$.
Note that $2\cdot 3=6=5+1\equiv 1\pmod{5}$. This means that in the ring $\mathbb{Z}_5$, the element $2$ is invertible and the inverse of $2$ (what you call $1/2$), is $3$.