Question

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

• A. an orthogonal matrix
• B. symmetric matrix
• C. skew-symmetric matrix
• D. none of these

Answer 1

Answer: (D)

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
$A=\left(\begin{array}{cc}1& 3\\ 0& 1\end{array}\right),A^{-1}=\left(\begin{array}{cc}1& -3\\ 0& 1\end{array}\right)$
Note that $B$ is an orthogonal matrix.
$C=A^{-1}BA=\left(\begin{array}{cc}1& -3\\ 0& 1<br/>\end{array}\right)\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\begin{array}{cc}1& 3\\ 0& 1\end{array}\right)$
$=\left(\begin{array}{cc}3& 10\\ -1& -3\end{array}\right)$
Note that $C$ is neither symmetric, nor skew-symmetric and nor-orthogonal.