Invertible Matrices over Finite Fields

If $p$ is prime, the number of $3\times3$ invertible matrices over $\mathbb{Z}_p$ is equal to the number of bases that can be chosen for the vector space $\mathbb{Z}_p^3$ (the three columns of the matrix are the elements of the basis).

The first basis element just has to be a nonzero vector, and there are $p^3-1$ of these. The second basis element has to be a vector that does not lie in the span of the first vector. The subspace spanned by the first vector is one-dimensional, and hence has $p$ elements. Thus there are $p^3 - p$ candidate vectors. Once we have chosen the first two basis vectors, the third vector must be a vector that lies outside the subspace spanned by the first two. The subspace spanned by the first two vectors is two-dimensional, and hence contains $p^2$ elements. Thus there are $p^3-p^2$ possible third basis vectors.

Thus the total number of basis vectors in $\mathbb{Z}_p^3$ , and therefore the total number of invertible $3\times3$ matrices over $\mathbb{Z}_p$ , is $(p^3-1)(p^3-p)(p^3-p^2)$ When $p=17$ , this is equal to $\boxed{111203278848}$ .