If A and Bare symmetric matrices and AB = BA, then $A^{-1}$B is a 

We have,

$\mathrm{AB}=\mathrm{BA}\Rightarrow {\mathrm{B}}^{\mathrm{\prime }}{\mathrm{A}}^{\mathrm{\prime }}=(\mathrm{AB}{)}^{\mathrm{\prime }}\Rightarrow \mathrm{AB}$ is symmetric 

Also, ${\mathrm{ABA}}^{-1}={\mathrm{BAA}}^{-1}=\mathrm{B}$

$\begin{array}{l}\Rightarrow {\mathrm{A}}^{-1}{\mathrm{ABA}}^{-1}-{\mathrm{A}}^{-1}\mathrm{B}\Rightarrow {\mathrm{BA}}^{-1}-{\mathrm{A}}^{-1}\mathrm{B}\\ \Rightarrow {\left({\mathrm{A}}^{-1}\mathrm{B}\right)}^{\mathrm{\prime }}={\left({\mathrm{BA}}^{-1}\right)}^{\mathrm{\prime }}\Rightarrow {\left({\mathrm{A}}^{-1}\right)}^{\mathrm{\prime }}{\mathrm{B}}^{\mathrm{\prime }}={\mathrm{A}}^{-1}\mathrm{B}\end{array}$

(Since, $A^{-1}$ is symmetric) 

So $A^{-1}$ B is a symmetric matrix.