Question. Let $\mathcal{M}_3$ be the set of $3\times 3$ matrices with non-negative integer entries and unit determinant. What is the number of $M\in \mathcal{M}_3$ with fixed sum of entries? What is the answer if we allow only matrices with strictly positive entries?

The rate of growth is also interesting. Below $n$ stands for the sum of entries.

Origin. For the similar question in the case of $2\times 2$ matrices we get $\phi(n)$ and $\phi(n) - 2$ respectively, see this recent post: Decomposition of a natural number as sum of positive integers.

Computation of the first elements. If we denote the answer sequence for non-negative case by $a_n$ then the first elements are

$a = [0, 0, 0, 3, 18, 54, 126, 261, 404, 667, 955, 1417, 1709, 2603, 2979, \ldots]$.

And if $b_n$ is the answer for the strictly positive case then $b_n = 0$ if $n \le 10$ and further we have

$b = [0,\ldots ,0, 9, 18, 49, 69, 169, 239, 388, 423, 848, 902, 1231, 1559, 2283, 2002, \ldots]$.

Unfortunately, both sequences do not seem to be in OEIS.

Thoughts on the growth. Evidently, $a_n\ge b_n$, and $a_n\le Cn^8$ because the number of partitions of $n$ into $9$ terms is $O(n^8)$. If we consider only upper-triangular matrices with ones on the diagonal, we will get the bound $a_n\ge Cn^2$.

Thus, the answer is of polynomial order. Note that this is different to the $2\times 2$ case where the number-theoretic properties played the main role.