Question

If $A=\frac{1}{3}\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]$ is an orthogonal matrix, then

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

Answer 1

Answer: (A), (C)

$a=-2$
$b=-1$

For an orthogonal matrix $A.A^T=A^T.A=I$

Therefore, $\frac{1}{3}\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]\times \frac{1}{3}\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow \frac{1}{9}\left[\begin{array}{ccc}9& 0& a+4+2b\\ 0& 9& 2a+2-2b\\ a+4+2b& 2a+2-2b& a^2+4+b^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Therefore, we get

$\frac{a+4+2b}{9}=0$

$\Rightarrow a+2b=-4$ ...$\left(i\right)$

$\frac{2a+2-2b}{9}=0$

$\Rightarrow 2a+2-2b=0$

$\Rightarrow a+1=b$ ...$\left(ii\right)$

and, $\frac{a^2+4+b^2}{9}=1$

$\Rightarrow a^2+b^2=5$ ... $\left(iii\right)$

On solving equation $\left(i\right),\left(ii\right)$ nd $\left(iii\right)$,we get

$a=-2,b=-1$ or

$a=1,b=2$

Hence, the correct options are $\left(A\right)$and $\left(C\right)$.