How to prove that the determinant of a symmetric matrix with the main diagonal elements zero and all other elements positive is not zero (i.e., that the matrix is invertible)?
EDIT: OP indicates in a comment that the entries above the diagonal are to be distinct.
Best Answer
\begin{vmatrix} 0 & 2 & 3 & 15 \\ 2 & 0 & 5 & 9 \\ 3 & 5 & 0 & 6 \\ 15 & 9 & 6 & 0 \end{vmatrix}
This matrix has determinant zero.
Added: The determinant in the $4\times 4$ case is quite nice: $$2[(af)^2+(be)^2+(cd)^2]-(af+be+cd)^2. $$ To get integer entries, consider the Diophantine equation $$2[x^2+y^2+z^2]=(x+y+z)^2 $$ which has solutions $x=cr^2$, $y=cs^2$, $z=ct^2$ where $r+s=t$. We need to be able to split $x,y,z$ into six distinct factors, two each. I eventually hit upon $c=3$, $r=2$, $s=3$, and $t=5$. This gives $x=12$, $y=27$, $z=75$ which I split as $a=2$, $f=6$, $b=3$, $e=9$, $c=15$, and $d=5$.