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What is the maximum possible determinant of a $6 \times 6$ matrix of $\pm1$?

This is the maximum I reached: $$\begin{vmatrix} 1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 \\ 1 & 1 & 1 & -1 & -1 & -1 \\ 1 & 1 & 1 & 1 & -1 & -1 \\ 1 & 1 & 1 & 1 & 1 & -1 \\ 1 & 1 & 1 & 1 & 1 & 1 \end{vmatrix} = 32$$

Mike Pierce
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Mahdi
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    [Close enough](http://math.stackexchange.com/questions/1465627/maximising-determinant-problem) – Jesse P Francis Nov 06 '15 at 06:34
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    http://mathworld.wolfram.com/HadamardsMaximumDeterminantProblem.html – Ivan Neretin Nov 06 '15 at 06:42
  • @IvanNeretin there is no 6*6 Hadamard matrix. – Mahdi Nov 06 '15 at 06:58
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    True, but still the link contains some relevant information. – Ivan Neretin Nov 06 '15 at 07:20
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    random check (with excel) gave $\left|\begin{matrix}-1 && 1 && -1 && 1 && 1 && -1 \\ -1 && -1 && 1 && -1 && -1 && 1 \\ -1 && -1 && -1 && 1 && 1 && 1 \\ -1 && 1 && 1 && -1 && 1 && 1 \\ 1 && -1 && 1 && 1 && 1 && -1 \\ -1 && 1 && 1 && 1 && -1 && 1\end{matrix}\right|=96$ (not sure if its maximum) – Jesse P Francis Nov 06 '15 at 07:48
  • Update: Brute Force: Found one with determinant $160$, I got a feeling that I've hit the roof! The logic I used is pretty sloppy, I can't assure if that's it. – Jesse P Francis Nov 06 '15 at 11:27
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    @JessePFrancis That is the best possible since it equals the [upper bound of Ehlich and Wojtas](https://en.wikipedia.org/wiki/Hadamard%27s_maximal_determinant_problem#The_Ehlich.E2.80.93Wojtas_bound_for_n.C2.A0.E2.89.A1.C2.A02_.28mod_4.29). – Will Orrick May 03 '16 at 15:10

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Another way to look at the problem: you can write your matrix $M$ (say) as $M=A+I$ where $A$ is a skew symmetric matrix and $I$ is a diagonal matrix. Then $|M|=|A|+|I|$ and see the determinant case of skew symmetric matrix when $n$ is even.

Kushal Bhuyan
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