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.

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.