If A is an invertible matrix and B is an orthogonal matrix. of the order same as that of A, then $C=A^{-1}BA$ is

Let $B=\left(\begin{array}{cc}\cos (\pi /2)& \sin (\pi /2)\\ -\sin (\pi /2)& \cos (\pi /2)\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$

and $A=\left(\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right),A^{-1}=\left(\begin{array}{cc}1& -3\\ 0& 1\end{array}\right)$

Note that B is an orthogonal matrix.

$\begin{array}{l}C=A^{-1}BA=\left(\begin{array}{cc}1& -3\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\begin{array}{ll}1& 3\\ 0& 1\end{array}\right)\\ =\left(\begin{array}{cc}3& 10\\ -1& -3\end{array}\right)\end{array}$

Note that C is neither symmetric, nor skew-symmetric and nor-orthogonal.