If $A=\frac{1}{3}\begin{bmatrix}1&2&2\\2&1&-2\\a&2&b\end{bmatrix}$ is an orthogonal matrix, then
Answer & explanation
Correct answer: option 2
Since A is an orthogonal matrix.
$∴AA^T = I$
$⇒\frac{1}{3}\begin{bmatrix}1&2&2\\2&1&-2\\a&2&b\end{bmatrix}.\frac{1}{3}\begin{bmatrix}1&2&a\\2&1&2\\2&-2&b\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$⇒\frac{1}{9}\begin{bmatrix}1&2&2\\2&1&-2\\a&2&b\end{bmatrix}\begin{bmatrix}1&2&a\\2&1&2\\2&-2&b\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$⇒\begin{bmatrix}9&0&a+4+2b\\0&9&2a+2-2b\\a+4+2b&2a+2-2b&a^2+4+b^2\end{bmatrix}=\begin{bmatrix}9&0&0\\0&9&0\\0&0&9\end{bmatrix}$
$⇒a+4+2b=0$ ...(1)
$2a+2-2b=0$ ...(2)
$a^2+4+b^2 = 9$ ...(3)
$\text{Eq. (2) - 2 × Eq. (1)}$
$-2-4b=0$
$⇒b=-1$
and, putting value of b in equation (1),
$a+4+(2)(-1)=0$
$a=-2$
$⇒a=-2, b=-1$