TITLE: How could we define the factorial of a matrix?
QUESTION [201 upvotes]: Suppose I have a square matrix $\mathsf{A}$ with $\det \mathsf{A}\neq 0$.
How could we define the following operation? $$\mathsf{A}!$$
Maybe we could make some simple example, admitted it makes any sense, with 
$$\mathsf{A} =
\left(\begin{matrix}
1 & 3 \\ 
2 & 1 
\end{matrix}
\right)
$$

REPLY [2 votes]: A factorial of an element of a group $n$ can be interpreted in an abstract algebraic sense, which can then be used with matrices. I will try to motivate the intuition here rather than ONLY giving a definition. Consider the expression $n!$ this is
$ 1 \times 2 \times 3 \times ... n  $
We now consider an arbitrary group $G$ with (not necessarily commutative operation +) then we can consider a particular element of $G$ (with generators $g_1, g_2 ... g_r$)  which we express minimally as the word
$$W = g_{\mu_1} + g_{\mu_2}+....  g_{\mu_s} $$
To be concrete if we let $G$ be the integers then there are two generators $1, -1$ and a positive integer number $N$ can be expressed in the way above as
$$N = N(1) = \underbrace{1+1+1...}_{N \ \text{times}} $$
Here $g_{\mu_k}$ is just positive $1$ the whole time.
If there was a a multiplication operation $\times$ defined now on this group as well (We haven't done it yet!) then we could define $W!$ as
$$ g_{\mu_1} \times \left( g_{\mu_1} + g_{\mu_2} \right) \times \ (g_{\mu_1} + g_{\mu_2} + g_{\mu_3} ) \times \ ...\times \  (g_{\mu_1} + g_{\mu_2} + ... g_{\mu_s})  $$
You can verify that if the group $G$ is the integers this results in
$$ 1 \times 2 \times 3 ... n $$
Which is what we would expect.
So the problem of defining a factorial on a group reduces very naturally to: "How to add a multiplication to a group"?
If $G$ is abelian then you can make $G$ into a ring (with possibly non commutative multiplication). If $G$ is not abelian then either a left or right near ring  is the way to go.
Once you add your multiplication then you get a natural factorial.