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If I recall correctly, most of the references out there, including Wolfram Alpha, define the inverse of a square matrix $A$ to be a square matrix $B$ of the same dimension such that $AB=I$. But is this really a sufficient condition? I mean, yes, if $A$ and $B$ are over commutative ring, the commutativity of $AB=BA$ comes for free, but what if they are not? Doesn't the above definition only define the $\textit{right}$ inverse? So one needs to define the $\textit{left}$ inverse as a square matrix $C$ such that $CA=I$?

Thanks.

user134070
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    When talking of square matrices it is enough to require one sided condition, since it automatically will give the other side. The proof I know is by means of groups (all square regular matrices over a field are a group under multiplication) – Timbuc Oct 16 '14 at 12:39

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