Does there exist a skew-matrix $A\in \mathbb{F}_3^{n\times n}$ with det$(A)\not=0$ and uneven $n$ ?
We can use det$(rA)=r^{n}$det$(A)$, for what it's worth.
I would be gratious about a hint in the right direction.
I would myself solve it with writing out the determinant, but that seems very tedious with unknown $n$.
$\det(A)=\det(A^T)=\det(-A)=(-1)^n \det(A)$.
If $n$ is odd this clearly implies $2\det(A)=0$, and in $\mathbb F_3$ this implies $\det(A)=0$.