to find the number of elments in $\vert\{A \in {F_4^{3\times 5}} \mid rk_{F_4}A=3 \}\vert$

Question

let for all $A \in {K^{m \times n}}$ $$\phi_A:K^{n\times 1}\rightarrow K^{m\times 1}$$ $$x \mapsto Ax$$ and $\phi_A$ ist a vectorspaceHomorphismus

and the Rang of $\phi_A$ is defined as $ Im_{\phi_A}=rk_{\phi_A}= rkA$

The question is to find the number of elemenets in $\{A \in {F_4^{5\times 3}} \mid rk_{F_4}A=3 \} $

$\textbf{My idea}$

let let for all $A \in {K^{5 \times 3}}$ $$\phi_A:F_4^{3\times 1}\rightarrow F_4^{5\times 1}$$ i do know that $dim({F_4^{3\times 1}})=kern_{\phi_A}+Im_{\phi_A}=3$

because of $Im_{\phi_A}=rk_{\phi_A}= rkA=3 \Rightarrow kern_{\phi_A}=0 \Rightarrow \phi_A $is injective function.

$\vert\{A \in {F_4^{5\times 3}} \mid rk_{F_4}A=3 \}\vert \Leftrightarrow \vert\{A \in {F_4^{5\times 3}} \mid \phi_A \text{is injective function.}\}\vert \Leftrightarrow \vert\{\phi_A \mid \phi_A \in MAP_{injective}({F_4^{3\times1}},{F_4^{5\times1}})\}\vert= \frac{\vert {F_4^{5\times1}}\vert!}{ {(\vert F_4^{5\times1}}\vert-\vert {F_4^{3\times1}}\vert)!}= \frac{4^5!}{(4^5-4^3)!}$

but the solution does give an other number which different than mine , can you please tell me where is my mistake , the right solution was $\textbf{1051807680}$

Answer 1

For $A\in {\mathbb{F}_4}^{3\times 5}$, it has rank 3 if and only if all three rows are linearly independent.

There are $4^5-1$ ways to select the first row, where minus 1 comes from the zero vector. The second row should not be a multiple of the first, so there are $4^5-4$ choices. The third row should not be any linear combination of the first and second row, there are $4^5-4^2$ possibilities for that.

Thus the desired number is $(4^5-1)(4^5-4)(4^5-4^2) = 1051807680$.