Question

If $A$ and $B$ are two non-singular matrices and both are symmetric and commute each other, then

• A. Both $A^{-1}B$ and $A^{-1}{B}^{-1}$ are symmetric
• B. $A^{-1}B$ is symmetric but $A^{-1}{B}^{-1}$ is not symmetric
• C. $A^{-1}{B}^{-1}$ is symmetric but $A^{-1}B$ is not symmetric
• D. Neither $A^{-1}B$ nor $A^{-1}{B}^{-1}$ are symmetric

Answer 1

The correct option is A Both $A^{-1}B$ and $A^{-1}{B}^{-1}$ are symmetric

$\begin{array}{c}\left|A\right|\ne 0\&\left|B\right|\ne 0\\ {\left(A^{-1}{B}^{-1}\right)}^T={\left(B^{-1}\right)}^T{\left(A^{-1}\right)}^T\\ {\left(A^{-1}{B}^{-1}\right)}^T={\left(B^T\right)}^{-1}{\left(A^T\right)}^{-1}\left[\begin{array}{c}{A}^T=A\\ B^T=B\end{array}\right]\\ {\left(A^{-1}{B}^{-1}\right)}^T=B^{-1}{A}^{-1}\\ {\left(A^{-1}{B}^{-1}\right)}^T={\left(AB\right)}^{-1}\left[AB=BA\right]\\ {\left(A^{-1}{B}^{-1}\right)}^T={\left(BA\right)}^{-1}\\ {\left(A^{-1}{B}^{-1}\right)}^T=A^{-1}{B}^{-1}\\ A^{-1}{B}^{-1}isasymmetricmatrix\end{array}$

similarly $A^{-1}B$ is also symmetric matrix.