Suppose we have $k$ matrices over $\mathbb{Z}_q$ of size $n \times n$, where $q \gg k$ and the entries of the matrices are chosen uniformly at random. Assume $q$ to be prime and we do not pick any matrix with all $0$ entries.
Can we obtain a bound on the probability that the sum of $k$ matrices is invertible?
Let us first forget about the restriction that none of the $k$ matrices $X_1, \dots, X_k$ is the all-zeros matrix.
Since we are working over $\mathbb{F}_q$, the sum $X = X_1 + \dots + X_k$ of $k$ matrices of size $n \times n$ whose entries are chosen uniformly at random, is a matrix with uniformly random entries. Equivalently, $X$ is a uniformly random element of $M_n(\mathbb{F}_q)$.
The probability that such an $X$ is invertible is well-studied, see for example: What is the number of invertible $n\times n$ matrices in $\operatorname{GL}_n(F)$?. Let's call this probability $p_{n, q}$.
You further specify that none of the matrices $X_1, \dots, X_k$ has all its entries be zero. The probability that some uniformly random $X_i$ is the all-zeros matrix is $\frac{1}{q^{n^2}}$. By the union bound inequality, the probability that any one of the $k$ matrices is the all-zeros matrix is at most $\frac{k}{q^{n^2}}$.
Thus when we do not pick all-zero matrices, the actual probability will be in the range $\left[p_{n,q} - \frac{k}{q^{n^2}}, p_{n, q} + \frac{k}{q^{n^2}}\right]$.
We are given that $q \gg k$, and thus $\frac{k}{q^{n^2}}$ seems like a small error term.
