<s>[INST] Number of $4\times3$ matrices of rank 3 over a field with 3 elements. I am finding number of $4\times3$ matrices of rank 3 over a field with 3 elements. If i count it as number of linearly independent columns i.e $3$ then its answer is $(3^{4}-1)(3^{4}-3)(3^{4}-3^{2}).$ But when i like to obtain the same formula as number of linearly independent rows my answer does not match. Please suggest me how to find the same formula as we look at number of linearly independent rows i.e. $3.$ Column wise already solved number of matrices of rank 3?. Thanks.
 [/INST] Start with the empty set and successively add the rows. The dimension of the space spanned by the set must increase from $0$ to $3$, and it can increase by at most $1$ in each step, so there must be exactly $3$ steps where it's incremented and one step where it stagnates. The incrementing steps always have the same counts of options, $3^3-3^0$, $3^3-3^1$ and $3^3-3^2$, whereas the stagnating step has $3^k$ options if it occurs after $k$ incrementing steps. Thus the total count is
$$
(3^3-3^0)(3^3-3^1)(3^3-3^2)(3^3+3^2+3^1+3^0)=(3^4-3^0)(3^4-3^1)(3^4-3^2)\;,
$$
in agreement with the column result. (For the equality, shift one factor of $3$ from $3^3-3^2$ to each of $3^3-3^0$ and $3^3-3^1$ and use $(3^1-3^0)(3^3+3^2+3^1+3^0)=3^4-3^0$.)
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