If $A$ is symmetric show that $(BA^{-1})^T(A^{-1}B^T)^{-1}=I$
I can see that:
$ (BA^{-1})^T(A^{-1}B^T)^{-1}\\ (A^{-1})^TB^T(B^T)^{-1}(A^{-1})^{-1}\\ A^{-1}B^T(B^T)^{-1}A\\ ...\\ A^{-1}...A=I $
I assume I would arrive at the last step, but I am confused how to get there.