Relation between determinant of  two matrices $A$ is a matrix with $1$ along diagonal and arbitrary unequal numbers less than $1$ on its non-diagonal. Let $B$ is a matrix with same $1$ along diagonal but having maximum non-diagonal element of $A$ as its non-diagonal element. What would be the relation between $\det (A)$ and $\det(B)$?  Suppose $A$ and $B$ are correlation matrices. Does increasing the off-diagonal entries i.e. correlation coefficient decreases determinant? 
 A: Enumerate the matrices $n \times n$ matrices $A_m$ with diagonal elements 1 and off-diagonal elements either $b$ or $0$, with at least one $0$ (there are $2^{n^2-n} - 1$ such matrices, but we can use symmetry to make the calculations somewhat more tractable).  For each $m$, $\det(A_m) = 1 - k b^2 + O(b^3)$  where $k$ is the number of pairs $(i,j)$ such that
$(A_m)_{ij} = (A_m)_{ji} = b$, while $\det(B) = 1 - \frac{n^2-n}{2} b^2 + O(b^3)$.  Thus $\det(A_m) \ge \det(B)$ for $0 \le b \le b_m$ where $ b_m > 0$. Then any $A$ with
diagonal elements 1 and off-diagonal elements in $[0,b]$ will have $\det(A) \ge \det(B)$ 
as long as $0 \le b \le \min_m b_m$.  I don't know if there's a closed-form formula for  $\epsilon(n) = \min_m b_m$.  The first few values, obtained by exhaustive enumeration, are $\epsilon(2) = \infty$, $\epsilon(3) = 1/2$, $\epsilon(4) = (4 - \sqrt{10})/3$, $\epsilon(5) = -3/4\,\cos \left( 1/3\,\arctan \left( 4/7\,\sqrt {2} \right)  \right) 
+5/4-3/4\,\sqrt {3}\sin \left( 1/3\,\arctan \left( 4/7\,\sqrt {2}
 \right)  \right)$.    This last is the least positive root of $-2+12 b-15 b^2+4 b^3$, corresponding to $A_m = \left[ \begin {array}{ccccc} 1&b&b&b&b\\ b&1&b&b&0\\ b&b&1&0&b\\ b&b&0&1&b
\\ b&0&b&b&1\end {array} \right]$
