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How $\det(\bf{A}+\bf{B})$ is related with $\det(\bf{A})$ where $\bf{A}$ is either semi-definite or positive definite matrix but $\bf{B}$ is a zero diagonal indefinite matrix. $\det(\cdot)$ denotes the determinant and all matrices are square. Off-diagonal elements of $\bf{B}$ are all positive. All elements of $\bf{A}$ are positive.

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    I appreciate this. First I was looking a general answer.@Gerry explained that some conditions are needed for one way inequality to be hold. For this reason, I have added conditions on the entries of elements. Thanks all for your kind considerations.2011-08-28

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$A=\pmatrix{2&-1\cr-1&2\cr}$ is positive definite with determinant 3. $B=\pmatrix{0&b\cr b&0\cr}$ has zero diagonal and is indefinite. The determinant of $A+B$ is $4-(b-1)^2$ which can be greater than, less than, or equal to 3. This suggests that in general you can't say very much.

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    All entries of $\bf{A}$ are positive and entries of $\bf{B}$ are non-negative.2011-08-28
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If $A \in \mathbb R^{n \times n}$ is positive definite, then it is invertible. Hence,

$\det (A+B) = \det (A (I_n + A^{-1} B)) = \det (A) \cdot \det (I_n + A^{-1} B)$

If $n=2$, then

$\det (I_n + A^{-1} B) = 1 + \det (A^{-1} B) + \mbox{tr} (A^{-1} B) = 1 + (\det (A))^{-1} \cdot \det (B) + \mbox{tr} (A^{-1} B)$

and, thus,

$\begin{array}{rl} \det (A+B) &= \det (A) \cdot \det (I_n + A^{-1} B)\\\\ &= \det (A) \cdot (1 + (\det (A))^{-1} \cdot \det (B) + \mbox{tr} (A^{-1} B))\\\\ &= \det (A) + \det (B) + \det (A) \cdot \mbox{tr} (A^{-1} B)\end{array}$

If $n > 2$, it gets a bit messy.