Dan Asimov wanted the NxN max-|determinant| sign matrix with equal numbers of +1's and -1's. (N necessarily even.) Some answers: N=2: maxdet=0. N=4: maxdet=16, use circulant(+---) then negate 2 rows. N=2^k for N>4: maxdet=N^(N/2) by Sylvester inductive doubling construction (use N/2 size Asimov solution Kronecker-producted with 2x2 Hadamard matrix) N=p+1 where p=prime and N divisible by 4: maxdet=N^(N/2). Use Paley construction of NxN Hadamard matrix by making pXp circulant, bordered by +1's in an extra first row and first column where the first row of the circulant is given by Jacobi symbols mod p, except falsely regard (0|p) as -1. Now negate the last N/4 columns, then finally negate the first row. N=6: maxdet=128. circulant(+-----) then negate 3 rows. This can be proven maximal by using http://www.indiana.edu/~maxdet/spectrum.html The first cases these constructions leave open now are N=10,14,18. You can probably solve these 3 cases but I doubt you will be able to settle this problem for all N. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)