Question

If the matrix $ \left[ \begin{matrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \\ \end{matrix} \right] $ is orthogonal, then

• A. $ \alpha =\pm \frac{1}{\sqrt{2}} $
• B. $ \beta =\pm \frac{1}{\sqrt{6}} $
• C. $ \gamma =\pm \frac{1}{\sqrt{3}} $
• D. All of these

Answer 1

Answer: (D)

Let $A=\left[\begin{array}{ccc}0 & 2 \beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & =\beta & \gamma\end{array}\right]$ Since, matrix $A$ is orthogonal $\therefore A A^{\prime}=1 \Rightarrow $
$$
\begin{array}{l}
{\left[\begin{array}{ccc}
0 & 2 \beta & \gamma \\
\alpha & \beta & -\gamma \\
\alpha & -\beta & \gamma
\end{array}\right]\left[\begin{array}{ccc}
0 & \alpha & \alpha \\
2 \beta & \beta & -\beta \\
\gamma & -\gamma & \gamma
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]} \\
\Rightarrow \left[\begin{array}{ccc}
4 \beta^{2}+\gamma^{2} & 2 \beta^{2}-\gamma^{2} & -2 \beta^{2}+\gamma^{2} \\
2 \beta^{2}-\gamma^{2} & \alpha^{2}+\beta^{2}+\gamma^{2} & \alpha^{2}-\beta^{2}-\gamma^{2} \\
-2 \beta^{2}+\gamma^{2} & \alpha^{2}-\beta^{2}-\gamma^{2} & \alpha^{2}+\beta^{2}+\gamma^{2}
\end{array}\right] \\
=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \text { Equating the corresponding }
\end{array}
$$
$\left[\begin{array}{ccc}0 & 2 \beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma\end{array}\right]\left[\begin{array}{ccc}0 & \alpha & \alpha \\ 2 \beta & \beta & -\beta \\ \gamma & -\gamma & \gamma\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ $\Rightarrow \left[\begin{array}{cc}4 \beta^{2}+\gamma^{2} & 2 \beta^{2}-\gamma^{2} & -2 \beta^{2}+\gamma^{2} \\ 2 \beta^{2}-\gamma^{2} & \alpha^{2}+\beta^{2}+\gamma^{2} & \alpha^{2}-\beta^{2}-\gamma^{2} \\ -2 \beta^{2}+\gamma^{2} & \alpha^{2}-\beta^{2}-\gamma^{2} & \alpha^{2}+\beta^{2}+\gamma^{2}\end{array}\right]$ $=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ Equating the corresponding elements of above matrices, we get $4 \beta^{2}+\gamma^{2}=1 ?$ (i) $2 \beta^{2}-\gamma^{2}=0 \ldots$ (ii) $\alpha^{2}+\beta^{2}+\gamma^{2}=1$ ? (iii) Adding Eqs. (i) and (ii), we get $6 \beta^{2}=1 \Rightarrow \beta=\pm \frac{1}{\sqrt{6}}$ From Eq. (ii), $\Rightarrow \gamma^{2}=2 \beta^{2} \Rightarrow \gamma^{2}=\frac{2}{6}=\frac{1}{3} \Rightarrow \gamma=\pm \frac{1}{\sqrt{3}}$
From Eq. (iii), $\alpha^{2}=1-\beta^{2}-\gamma^{2} \Rightarrow $
$\alpha^{2}=1-\frac{1}{6}-\frac{1}{3}=\frac{1}{2} \Rightarrow \alpha=\pm \frac{1}{\sqrt{2}}$