Question

Let $A=\left[\begin{array}{ccc}0& 2b& c\\ a& b& -c\\ a& -b& c\end{array}\right]$ be an orthogonal matrix , then the values of $a,b,c$ are

• A. $b=\pm \frac{1}{\sqrt{6}},c=\pm \frac{1}{\sqrt{3}}$
• B. $a=\pm \frac{1}{\sqrt{2}},c=\pm \frac{1}{\sqrt{6}}$
• C. $a=\pm \frac{1}{\sqrt{2}},b=\pm \frac{1}{\sqrt{6}}$
• D. all of these

Answer 1

The correct option is D all of these

Let $A=\left[\begin{array}{ccc}0& 2b& c\\ a& b& -c\\ a& -b& c\end{array}\right]$ and $A^T=\left[\begin{array}{ccc}0& a& a\\ 2b& b& -b\\ c& -c& c\end{array}\right]$

As $A$ is orthogonal $\therefore A{A}^T=I$

$\Rightarrow \left[\begin{array}{ccc}0& 2b& c\\ a& b& -c\\ a& -b& c\end{array}\right]\left[\begin{array}{ccc}0& a& a\\ 2b& b& -b\\ c& -c& c\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

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

$\therefore$ Using definition of equality of two matrices

$4b^2+c^2=1,2b^2-c^2=0,a^2+b^2+c^2=1$

On solving them, $a=\pm \frac{1}{\sqrt{2}},b=\pm \frac{1}{\sqrt{6}},c=\pm \frac{1}{\sqrt{3}}$