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Let $A$ and $B$ be $n\times n$ matrices over $\mathbb R$, with $n\ge 4$. $A$ is a symmetric matrix with inverse $A^{-1}$ and $B$ is an antisymmetric matrix.

Consider $$ f= \left[\det (A+B) \right]^{1/2} $$ where $\det$ is the matrix determinant and we can assume that $\det (A+B)> 0$. I want to find an expression for the contributions to $f$ that are quadratic in the matrix elements $B_{ab}$ of $B$.

I believe it must be of the form $\alpha B_{ab} (A^{-1})^{ac}(A^{-1})^{bd}B_{cd}= -\alpha \, \mathrm{tr}\, (B A^{-1}B A^{-1})$ where tr is the standard matrix trace and $\alpha$ is some coefficient.

Unfortunately I have no idea how to show this in a simple way. Brute force calculation might help, but I don't know how I would have to handle the square root.

Any help on this will be much appreciated.

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Turns out the answer is pretty simple. The symmetry properties of $A$ and $B$ are irrelevant. Just use the expansion $$ \det (\mathbb 1 + A) = 1 + \mathrm{Tr}\,{A} + \frac{1}{2} \left(\mathrm{Tr}\,{A}\right)^2 - \frac{1}{2} \mathrm{Tr}\,{A^2} + o(A^3) $$ and then expand the square root. The terms with $\left(\mathrm{Tr}\,{A}\right)^2$ cancel out and the term $\mathrm{Tr}\,{A^2}$ has a factor $-1/4$.

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