Question

If $A = \frac{1}{3} \begin{bmatrix}1&2&2\\ 2&1&-2\\ a&2&b\end{bmatrix} $ is an orthogonal matrix, then

• A. a = - 2, b = - 1
• B. a = 2, b = 1
• C. a = 2, b = - 1
• D. a = - 2, b = 1

Answer 1

Answer: (A)

As A is an orthogonal matrix, $AA^T = I$
$\Rightarrow \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} $
$\Rightarrow \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} $
$\Rightarrow \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} $
$\Rightarrow a + 4 + 2b = 0, 2a + 2 - 2b = 0 , a^{2} + 4 + b^{2} = 9 $
$\Rightarrow a + 2b + 4 = 0, a - b + 1 = 0 a^{2} + b^{2} = 5 $
$\Rightarrow a = - 2, b = - 1$