Let $R$ be a unital (but not necessarily commutative) ring, $M$ a finitely generated left $R$-module, and $f$ an endomorphism of $M$. Fix a set of generators $g_1,\ldots,g_n$ and say a matrix $C=(c_{ij}) \in M_n(R)$ represents $f$ if the $i$-th row of $C$ is a coordinate vector for $f(g_i)$, i.e. if $\sum_{j=1}^n c_{ij}g_j = f(g_i)$. If we take any element $x\in M$, written with our generators as $x=r_1g_1+\cdots +r_ng_n$, then $$ (r_1,\ldots,r_n)C $$ is a coordinate vector for $f(x)$. If $M$ is free and $g_1,\ldots,g_n$ is a basis for $M$, then $C$ is uniquely determined, and $f$ is invertible if and only if $C$ is.
I am interested in the case when $M$ is not free. Then $C$ is no longer uniquely determined and not every matrix represents an endomorphism. I want to know what can be said about the relationship between the invertibility of $C$ and the invertibility of $f$. It can happen that $f$ is invertible but $C$ is not. It can also happen that no matrix representing $f$ is invertible (for a fixed set of generators). In the commutative case, if $C$ is invertible, then $f$ is as well and $C^{-1}$ represents $f^{-1}$. What about if $R$ is not commutative? Also, if $f$ is invertible, does there always exist a generating set $g_1,\ldots,g_n$ and a matrix $C$ representing $f$ such that $C$ is invertible? The answer is yes if $|M|$ is finite, or if $M$ is projective. But I don't know if it's always the case.
Does there exist any literature about this matrix representation of endomorphisms of f.g. modules?