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Recently, I came across an interesting problem:

Consider a matrix $A\in M(3\times 3)$ whose entries are (pairwise different) prime numbers. What values $\operatorname{rank}(A)$ can take?

At first, I thought it must be that $\operatorname{rank}(A) = 3$. However, a bit of computation proved that my intuition was incorrect as the rank can be lower $$ \operatorname{rank}\begin{pmatrix} 5 & 7 & 11\\ 17 & 19 & 23 \\ 41 & 43 & 47 \end{pmatrix} = 2.$$

Although the problem is technically solved, I'm curious if there are some additional conditions on entries under which such matrices are of maximal rank.

I attacked the problem with Bézout's identity and managed to obtain some identities, but its rather messy and I don't like it at all. My questions would be then: (1) is there any reasonable answer to this problem? (2) what happens in case $A\in M(n\times n)$ when $n>3$?

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This is probably overkill, but the Green-Tao Theorem guarantees that you can get rank-two for any $n$: take an arithmetic progression of length $n^2$ and the difference between rows will be constant. Using arithmetic progressions of different step you should then achieve any rank

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For matrices in $M(n\times n)$ Rank 2 can be attained by using the Green-Tao Theorem referenced by @Martin Argerami.

Let the elements of the matrix be the terms of the AP entered sequentially from left to right and from top top to bottom. Subtracting each Row $i$ from Row $i+1$, for $1\le i<n$, reduces all elements outside the top row to $n$ times the common difference and so the matrix has Rank 2.

Rank 1 cannot be attained as per the comment by @John Omielan.

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