Finding Maximum Determinant of a $6\times 6$ Matrix

Question

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

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

Answer 1

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.