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. $a=-2,b=-1$
• 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,
$A{A}^T=I$
$\Rightarrow \frac{1}{3}\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right].\frac{1}{3}\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow \frac{1}{9}\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right].\frac{1}{3}\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{ccc}9& 0& a+4+2b\\ 0& 9& 2a+2-2b\\ a+4+2b& 2a+2-2b& a^2+4+b^2\end{array}\right]$
$=\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]$
$\Rightarrow a+4+2b=0,2a+2-2b=0$ and $a^2+4+b^2=9$
$\Rightarrow a+2b+4=0,a-b+1=0$ and $a^2+b^2=5$
$\Rightarrow a=-2,b=-1$