I want to show that $SL_2(\bf F_3)$ is generated by $A=\begin{bmatrix} 1\ 1 \\ 0\ 1 \end{bmatrix}$ and $B=\begin{bmatrix} 1\ 0 \\ 1\ 1 \end{bmatrix}$, using the fact that the order of this group is 24. Since these matrices have the determinant 1, every matrix generated by these two matrices has the det 1, so it is enough to show that there are 24 distinct matrices in $\langle A, B\rangle$. I found that $|A|=|B|=3$ and $|AB|=|BA|=4$. So there are 4 elements of order 3, 4 elements of order 2, 2 elements of order 4 and 1 elements of order 1. But in this way, there's no assurance that every element will be found and it is too tedious. Is there an elegant way to prove?
$\bf Add)$ Is $SL_2(\bf F_3)$isomorphic to $S_4$?
You know your matrices generate a subgroup of $H \leq \mathrm{SL}_2(3)$. Thus $|H|$ divides $24$. $H$ has at least 2 Sylow-3-subgroups (generated by $A$ and $B$ respectively) hence there are by Sylow's theorem at least $4$ Sylow-3-subgroups. This yields $8$ elements of order $3$. $AB, (AB)^3,BA,(BA)^3$ are $4$ distinct elements of order $4$. So together with the unit element $H$ contains at least $13$ elements. As a subgroup of a group with $24$ elements $H$ is therefore equal to $\mathrm{SL}_2(3)$.