Question

If $A=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 2\\ 2& 1& -2\\ x& 2& y\end{array}\right]$ is orthogonal, then x + y =
(a) 3
(b) 0
(c) − 3
(d) 1

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},A=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 2\\ 2& 1& -2\\ x& 2& y\end{array}\right] \\ \Rightarrow A^T=\frac{1}{3}\left[\begin{array}{ccc}1& 2& x\\ 1& 1& 2\\ 2& -2& y\end{array}\right] \\ \mathrm{Now},A^{T}A=I \\ \Rightarrow \left[\begin{array}{ccc}{x}^2+5& 2x+3& xy-2\\ 3+2x& 6& 2y\\ xy-6& 2y& y^2+8\end{array}\right]=\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right] \\ \mathrm{The}\mathrm{corresponding}\mathrm{elements}\mathrm{of}\mathrm{two}\mathrm{equal}\mathrm{matrices}\mathrm{are}\mathrm{not}\mathrm{equal}. \\ \mathrm{Thus},\mathrm{the}\mathrm{matrix}A\mathrm{is}\mathrm{not}\mathrm{orhtogonal}. \end{gathered}$$