Question

If \[A = \frac{1}{3}\begin{bmatrix}1 & 1 & 2 \\ 2 & 1 & - 2 \\ x & 2 & y\end{bmatrix}\] is orthogonal, then x + y =

(a) 3
(b) 0
(c) − 3
(d) 1

पर्याय

• 3

• 0

• -3

• 1

• None of these

Answer 1

None of these

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