# a few things to possibly add

* $|SL(n,q)| = \frac{1}{q-1}\prod_{i=0}^{n-1}(q^n-q^i)$ where q is the order of the field.

* $[SL(n,q),SL(n,q)] = SL(n,K)$ with two exceptions:
$SL(2,2)$ and $SL(2,3)$ where q is the order of the field.

Parting words from the person who closed the correction:
Just like for GL, I think these results would be great to add into a "$\operatorname{SL}(n,\mathbb{F}_q)$" entry, but are of too narrow interest for SL in general.
Status: Rejected
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What kind of correction is this:

### derivative of a determinant

The determinant of a matrix $A(t)=A_{ij}(t)$ is given by

 $det(A_{ij}(t))=e_{ijk}A_{i1}(t)A_{j2}(t)A_{k3}(t),$

where $e_{ijk}$ is the Cartesian alternator. Let us take the derivative of this expression with respect to the parameter $t$,

 $\frac{d}{dt}det(A_{ij}(t))=e_{ijk}(A_{i1}(t),tA_{j2}(t)A_{k3}(t)+A_{i1}(t)A_{j% 2}(t),tA_{k3}(t)+A_{i1}(t)A_{j2}(t)A_{k3}(t),t),$

where “$,t$” stands for derivation. Let us now consider the quantity

 $C_{k3}(t)=e_{ijk}A_{i1}(t)A_{j2}(t)=\frac{1}{2}(e_{ijk}A_{i1}(t)A_{j2}(t)+e_{% jik}A_{j1}(t)A_{i2}(t))=\frac{1}{2}e_{ijk}(A_{i1}(t)A_{j2}(t)-A_{i2}(t)A_{j1}(% t)),$

which is the same as

 $C_{k3}(t)=\frac{1}{2}e_{ijk}e_{lm3}A_{il}(t)A_{jm}(t),$

and in general,

 $C_{kn}(t)=\frac{1}{2}e_{ijk}e_{lmn}A_{il}(t)A_{jm}(t).$

This is the so-called “adjugate” $adj$ of the matrix $A(t)$ (notice that this formula is also useful to find the inverse of $A(t)$ whenever it be no singular). Thus, the sum for $\frac{d}{dt}det(A_{ij}(t))$ above, may be expressed by

 $\frac{d}{dt}det(A_{ij}(t))=\frac{1}{2}e_{ijk}e_{lmn}A_{il}(t)A_{jm}(t)A_{kn}(t% ),t=C_{kn}(t)A_{kn}(t),t,$

which is clearly the “trace” $tr$ of this product. So that, in matrix notation we have

 $\frac{d(detA(t))}{dt}=tr(adj(A(t))\frac{d(A(t))}{dt}).$

Indeed this formula is also valid for any finite dimension $n$, and its proof is a little more elaborated. We have exposed here the case for $n=3$ for reason of brevity. Also note that $C_{kn}^{T}=C_{nk}$ is a cofactor of the matrix $A(t).$