We give some necessary conditions for maximality of $0/1$-determinant. Let ${\bf M}$ be a nondegenerate $0/1$-matrix of order $n$. Denote by $\bf A$ the matrix of order $n+1$ which appears from ${\bf M}$ after adding the $(n+1)$th row $(0,0,\ldots,0,1)$ and the $(n+1)$th column consisting of $1$'s. Suppose ${\bf A}^{-1}=(l_{ij}),$ then for all $i=1,\ldots,n$ we have $\sum_{j=1}^{n+1} |l_{ij}|\geq 2.$ Moreover, if $|\det({\bf M})|$ is equal to the maximum value of a $0/1$-determinant of order $n$, then $\sum_{j=1}^{n+1} |l_{ij}|= 2$ for all $i=1,\ldots,n$. Keywords: maximum 0/1-deteminant, simplex, cube, axial diameter