Number of encrypting $n\times n$ matrices modulo 26

Question

I'm trying to find the number of $n$ x $n$ encrypting matrices modulo 26 for $n$ $\in$ $\Bbb N$.

I really don't know what method to use to approach this... does it involve Hill ciphers?

Answer 1

The number of $n\times n$ invertible matrices modulo a prime $p$ is:

$$(p^n-1)(p^n-p)\cdots(p^n-p^{n-1})=p^{n(n-1)/2}(p-1)(p^2-1)...(p^n-1)$$

The number of invertible matrices modulo $26$ is the product of the numbers for $p=2$ and $p=13$. (This is an application of the Chinese remainder theorem to matrices.)

So that gives you:

$$26^{n(n-1)/2} (13^1-1)(2^1-1)\cdot(13^2-1)(2^2-1)\cdot(13^{n-1}-1)(2^{n-1}-1)$$