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I'm in a linear algebra course, working with Hoffman, and the professor asked me if there is a way to count $n\times n$ matrix over a finite field $\mathbb{Z}_p$. Then give us the next question: let $M$ be a $3\times 3$ matrix over the field $\mathbb{Z}_2=\{0,1\}$ how many matrices with $|M| = 1$ are there?

user326159
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Gustavo Adolfo Inciarte Briceo
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  • One quick note is that this is exactly the same as asking how many matrices over $\mathbb{Z}/2\mathbb{Z}$ are invertible, because the only possible nonzero determinant is $1$, or equivalently the $3\times 3$ matrices over $\mathbb{Z}/2\mathbb{Z}$ which are of full rank. – walkar Jun 01 '24 at 14:32
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    This is something your professor wanted YOU to find out, so why did you not try yourself? – trula Jun 01 '24 at 14:38
  • So there are 3 matrices ? I'm trying this by myself since a week ago, the only biography I have is Hoffman, so I'm looking for a way to count it, if there exists – Gustavo Adolfo Inciarte Briceo Jun 01 '24 at 14:48

1 Answers1

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The group $SL(n,q)$ of $n\times n$-matrices over $\Bbb F_q$ has exactly

$$ \frac{(q^n-1)(q^n-q)(q^n-q^2)\cdots (q^n-q^{n-1})}{q-1} $$

many elements, see this post:

Order of general- and special linear groups over finite fields.

For $q=2$ this number is $(2^3-1)(2^3-2)(2^3-2^2)=7\cdot 6\cdot 4=168$. So there are $168$ matrices of size $3$, with determinant $1$, because this is the order of the group $SL(3,2)$, which consists exactly of all $3\times 3$-matrices over $\Bbb F_2$ with determinant $1$.

Dietrich Burde
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