Which is the max value of the determinant with 4 lines and 4 colums ,where every term is +- 1?

Question

I understand that this problem can be solve with the volume of a tetrahedron.

But i don't know how.

please help me !

\begin{vmatrix} \pm1 & \pm1 & \pm1 & \pm1 \\ \pm1 & \pm1 & \pm1 & \pm1 \\ \pm1 & \pm1 & \pm1 & \pm1 \\ \pm1 & \pm1 & \pm1 & \pm1 \notag \end{vmatrix}

Answer 1

You have $|\det[x_1 \cdots x_n ] | \le \|x_1\|\cdots \|x_n\|$ (Hadamard inequality).

This shows the determinant is upper bounded by 16.

Now choose $A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}$, and compute $\det A = 16$.