Largest determinant of a real $3\times 3$-matrix

Question

What is the largest determinant of a real $3\times 3$-matrix with entries from the interval $[-1,1]$ ? A result of John Williamson says that the largest value is equal to $4$, if the entries are just either $1$ or $-1$. Is this still true for all values in $[-1,1]$ ? For complex matrices with entries $|a_{ij}|\le 1$ it is not true. We have $|\det(A)|\le 3\sqrt{3}$, and equality can be attained with a Vandermonde type of matrix containing the third roots of unity.

Answer 1

The determinant is (up to sign and a factor of $6$) the volume of a tetrahedron having the origin and the three column vectors as vertices. Moving one vertex further from the plane determined by the other three vertices increases this volume. By convexity of the cube, among the farthest points is at least one vertex. Thus the result for $[-1,1]$ is the same as for $\{-1,1\}$.