Requirements on elements for the inversion of matrices to be well defined. I'm trying to learn some algebra here. Taken a few more or less clumsy steps on groups and in some fields.
If I have a matrix with elements from some set, what are the requirements on the set and the operations addition and multiplication to guarantee that matrix inverse exists in some sense? 
Own work Well at least some thoughts I have had...


*

*Addition and multiplication must be well defined on the elements, or how else would we define matrix multiplication in the first place? It is usually defined recursively on the elements as sums of products, right..?

*Additive and multiplicative identity elements should be necessary - or what else should "0" and "1" in the $I_n$ mean?

*Likewise I guess additive and multiplicative inverses should exist, but I lack some arguments for it.

*Is associativity necessary? Also here I lack arguments for or against.

 A: For addition to be defined, it helps for the matrices to be "the same size" (both $m \times n$, for example). It's hard to decide what $a_{ij} + \emptyset$ might mean.
For multiplication to be defined, it likewise is helpful for the matrices to be "square" (for example, $n\times n$), because only matrices of matching row-versus-column sizes can be multiplied in the "usual" way.
In order for the usual matrix multiplication to be carried out, and for there to be an identity matrix, and for matrix multiplication to be associative, and distributive over addition, a good starting place is thus a ring $R$ with unity.
While it is perfectly possible to define matrices over such a ring, the bookkeeping required gets cumbersome, and so most texts limit the ring to be commutative (which simplifies the calculation of matrix products considerably).
In doing so, one obtains what is called an $R$-algebra of $n \times n$ matrices over the ring $R$. Let's call this ring $M_n(R)$ (for matrix ring of "size" $n$). What your question is, then, is: what is the group of units of $M_n(R)$?
The thing is, if $R$ is a commutative ring with unity, and $n > 1$, it's always possible to create zero-divisors in $M_n(R)$, for example with $n = 2$ we have:
$\begin{bmatrix}0&1\\0&0\end{bmatrix}$ (among others),
and such a matrix cannot be invertible. However, with $n = 1$, we can ensure a non-zero $1 \times 1$ matrix is invertible, by requiring $R$ to be a field (or a division ring, if we drop the commutative stipulation).
Note that since $M_n(R)$ for $n > 2$ contains zero-divisors, we cannot "make inverses" by simply "enlarging our ring" (like we can do when we make fractions from integers). This is because, for nice enough rings $R$, the ring $M_n(R)$ models the set of $R$-linear functions from $R^n \to R^n$, and it should be clear upon reflection, that such a function need not be bijective.
By and large, the "usual" approach is just to limit our study at the outset to invertible matrices, which, although they are not closed under matrix addition, ARE closed under the operation of matrix multiplication, thus yielding a group.
