If A and B are symmetric non-singular matrices, AB = BA and then

Since A and B are symmetric, $A^T = A and B^T = B$

$AB=BA\Rightarrow {\left(AB\right)}^{-1}={\left(BA\right)}^{-1}\Rightarrow B^{-1}{A}^{-1}=A^{-1}{B}^{-1}$

${\left(A^{-1}B\right)}^T=B^T{\left(A^{-1}\right)}^T=B{\left(A^T\right)}^{-1}=B{A}^{-1}$

$B{A}^{-1}=A^{-1}A.B{A}^{-1}=A^{-1}B.A{A}^{-1}=A^{-1}BI=A^{-1}B$

$\therefore A^{-1}B\text{ is symmetric}$

$\text{Similarly b and c can be proved}$