Mathematics - Matrix Question with Solution | TestHub

We know that, if $A$ is an orthogonal matrix, then $\begin{array}{c}\Rightarrow A{A}^T=I\\ \Rightarrow \\ \Rightarrow \left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]\cdot \frac{1}{3}\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ \Rightarrow \left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]=\left[\begin{array}{lll}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]\\ \left[\begin{array}{ccc}9& 0& a+4+2b\\ 0+4+2b& 9& 2a+2-2b\\ 2a+2-2b& a^2+4+b^2& \end{array}\right]\\ =\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]\\ \Rightarrow a+4+2b=0\text{ and }2a+2-2b=0\\ a+2b=-4\dots (i)\\ a-b=-1\dots (ii)\end{array}$ By solving Eqs. (i) and (ii), we get $a=-2,b=-1$