I've been checking the following discrete mathematics exercise:
How many 2x2 matrices with module 27 inputs exist with the condition being invertible?
Checking many information in the internet I found that one answer proposed is $526178$, another one is $314928$ but through different calculations I've been unable to find those numbers.
Any help will be really appreciated
On
Idea: The number of all $2\times 2$ matrices is $27^4$. Calculate the number of matrices that are not invertibile, so $$27\mid ad-bc$$ if $a,b,c,d\in \mathbb{Z}_{27}$ and
$$ M= \left[ \begin{matrix} a&b\\ c&d\\ \end{matrix} \right] $$
Divide it on cases
If $3\nmid d$ then $a$ is uniqely determined with $b,c,d$ with formula $a= bcd^{-1}{\mod 27}$
If $3\mid d$ and $9\nmid d$ ...
If $9\mid d$ and $27\nmid d$ ...
If $27\mid d$ ...
Let the matrix have rows $a, b$ and $c, d$.
If the matrix is invertible mod $27$, then no row or column can be a vector multiple of $3$. There are two cases:
Case 1: If $3\mid a$ then neither $b$ or $c$ can be a multiple of $3.$ There are $9$ choices for $a$ and $18$ for each of $b$ and $c$. For a given choice of $a, b,$ and $c$, we need to exclude any values for $d$ that makes $ad-bc$ divisible by $3$. But $bc$ is not divisible by $3$ while $a$ is. So the congruence $ad\equiv bc \pmod{3}$ has no solutions. Therefore $d$ can be any of the $27$ numbers. We have $9\cdot 18\cdot 18\cdot 27 = 78732$ possibilities for Case 1.
Case 2: If $3\nmid a$, then $b$ and $c$ can be anything. So we have $18$ choices for $a$ and $27$ each for $b$ and $c$. Since $3\nmid a$, the congruence $ax=cd \pmod{3}$ has a unique solution which lifts to $9$ solutions $\pmod{27}.$ These must be excluded, so there are $18$ choices for $d$. We have $18\cdot 27 \cdot 27\cdot18 = 236196$ possibilities for Case 2.
We add the two cases and get $314928$ for the final answer.