If A,B are symmetric matrices, then prove that (BA^{-1})^T(A^{-1}B^T)^{-1} = I

If A,B are symmetric matrices, then prove
that $\left(B{A}^{-1}{\right)}^{T}\left({A}^{-1}{B}^{T}{\right)}^{-1}=I$
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Step 1
Given that, A,B are symmetric matrices.
Prove that $\left(B{A}^{-1}{\right)}^{T}\left({A}^{-1}{B}^{T}{\right)}^{-1}=I$
Step 2
Consider the LHS,
$\left(B{A}^{-1}{\right)}^{T}\left({A}^{-1}{B}^{T}{\right)}^{-1}=\left({A}^{-1}{\right)}^{T}\left(B{\right)}^{T}\left({B}^{T}{\right)}^{-1}\left({A}^{-1}{\right)}^{-1}$

$=RHS$
Hence , the required is obtained.

Jeffrey Jordon