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How to find $|\{A \in \mathbb F _7^{5 \times 5}|A \text{ is invertible}\}|$?

$$ \begin{align} GL_n(K) & = \{ A \in K^{n \times n }| A \text{ is invertible}\} \\ & = \{ A \in K^{n \times n }| \phi_A \text{ is an isomorphism}\} \\ & = \{ A \in K^{n \times n }| rk_{\phi A} = \text{dim }K^{n \times n}\} \\ & = \{ A \in K^{n \times n }| rk A = n\} \end{align} $$

And also

$Col(A) = Im \phi A$

$rk_KA = dim_kCol(A)$

Question: How can I zip this information to get the cardinality of all possible invertible matrices in $\mathbb F _7^{5 \times 5}$?

This post seems equivalent but somehow it does not help me. A technique without determinants would be the best for me at the moment. I appreciate also solutions or help using determinants.

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    $\begingroup$ Take a look at the answers here or here. At least some of the answers are generic. I would close this as a duplicate, but need to ask more experienced moderators about what that will do to the bounty. $\endgroup$ – Jyrki Lahtonen Jun 11 '17 at 20:59
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The columns must form a basis of $\Bbb F_7^5$. For the $k$th column, you can pick any of $7^5$ vectors, except those $7^{k-1}$ that are in the $(k-1)$-dimensional subspace generated by the preceding $k-1$ columns. So the count is $$(7^5-7^0)(7^5-7^1)(7^5-7^2)(7^5-7^3)(7^5-7^4) $$

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  • $\begingroup$ Unfortunately I do not fully understand you explanation. Why do you subtract $7^0, 7^1, 7^2...$? Yes, you mention it has something to do with the $(k-1)$-dimensional subpace.. $\endgroup$ – jublikon Jun 7 '17 at 7:15
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    $\begingroup$ He counts the numer of matrices as follows. First he observe that any martix is given by 5 vectors (of lenth 5). A matrix is invertible if and only if these vectors are linearily independent. Now let's count for the first vector we have no restrictions except from not being 0, that is $7^5-1$. The second vector can't be in the span of the first vector, this span contains 7 elements so we have $7^5-7$. Do this 3 more times and you're done. $\endgroup$ – Yanko Jun 11 '17 at 21:01

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