Question

$A=\frac{1}{3}\left(\begin{array}{ccc}1& -2& 2\\ -2& 1& 2\\ -2& -2& -1\end{array}\right)$ is

• A. orthogonal
• B. involutory
• C. symmetric
• D. diagonal

Answer 1

The correct option is A orthogonal

Given that
$A=\frac{1}{3}\left(\begin{array}{ccc}1& -2& 2\\ -2& 1& 2\\ -2& -2& -1\end{array}\right)$
$A^{\prime }=\frac{1}{3}\left(\begin{array}{ccc}1& -2& -2\\ -2& 1& -2\\ 2& 2& 1\end{array}\right)$
$\therefore A{A}^{\prime }=\frac{1}{3}\left(\begin{array}{ccc}1& -2& 2\\ -2& 1& 2\\ -2& -2& -1\end{array}\right)\times \frac{1}{3}\left(\begin{array}{ccc}1& -2& -2\\ -2& 1& -2\\ 2& 2& 1\end{array}\right)$
$=\frac{1}{9}\left(\begin{array}{ccc}1+4+4& -2-2+4& -2+4-2\\ -2-2+4& 4+1+4& 4-2-2\\ -2+4-2& 4-2-2& 4+4+1\end{array}\right)$
$=\frac{1}{9}\left(\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=I$
Hence $A$ is orthogonal matrix.