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Inspired by a previous question, given a square non-symmetric matrix whose elements are all prime but distinct from each other, does this guarantee that the matrix is invertible? It's easy to see $N=2$ this holds, a counter-example would imply that there must be four distinct primes such that $p_1 p_2 = p_3 p_4$.

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No. To find a $3 \times 3$ counterexample it suffices to write down three $3$-term prime arithmetic progressions with the same common difference $d$. By inspection $\bmod 2,3$ the smallest possible such difference is $d = 6$, which gives $$\left[ \begin{array}{ccc} 5 & 11 & 17 \\ 7 & 13 & 19 \\ 31 & 37 & 43 \end{array} \right].$$

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I suppose it is not even necessary for the common differences to be the same, but the next possible common difference is $d = 12$ which I think would lead to a bigger example. –  Qiaochu Yuan Aug 10 '12 at 13:41
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A $4 \times 4$ counterexample is $$\pmatrix{43 & 17 & 53 & 19 \cr 11 & 71 & 3 & 89 \cr 79 & 47 & 29 & 37 \cr 31 & 7 & 61 & 13 \cr }$$

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Thank you for the 4x4 counterexample Robert. Was there any special construction that you used to find it like Qiaochu? –  Hooked Aug 10 '12 at 3:41
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Nothing so clever. Actually I did a random search among permutations of the primes $< 100$ taken $16$ at a time. –  Robert Israel Aug 10 '12 at 6:33
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