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 (BA^{-1})^T(A^{-1}B^T)^{-1} = I

Question
Matrices
asked 2020-12-30
If A,B are symmetric matrices, then prove
that \((BA^{-1})^T(A^{-1}B^T)^{-1} = I\)

Answers (1)

2020-12-31
Step 1
Given that, A,B are symmetric matrices.
Prove that \((BA^{-1})^T(A^{-1}B^T)^{-1} = I\)
Step 2
Consider the LHS,
\((BA^{-1})^T(A^{-1}B^T)^{-1} = (A^{-1})^T(B)^T(B^T)^{-1}(A^{-1})^{-1}\)
\(=(A^{-1})^T IA \ \ \ \ \left[ \because (B^T)(B^T)^{-1}=I \text{ and } (A^{-1})^{-1}=A \right]\)
\(=(A^{-1})^T A^T \ \ \ \ \left[ \because A=A^T \right]\)
\(=(A^{T})^{-1} A^T \ \ \ \ \left[ \because (A^T)^{-1}=(A^{-1})^T \right]\)
\(=I \ \ \ \ \left[ \because (A^T)(A^T)^{-1}=I \right]
\(=RHS\)
Hence , the required is obtained.
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