Question

Determine the values of $\alpha ,\beta ,\gamma$ when
$\left[\begin{array}{ccc}0& 2\alpha & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right]$ is orthogonal.
Find the value of $1/|\alpha \beta \gamma |$

Answer 1

Let $A=\left[\begin{array}{ccc}0& 2\alpha & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right]$
$\therefore A^{\prime }=\left[\begin{array}{ccc}0& \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{array}\right]$
But given $A$ is orthogonal
$\therefore A{A}^{\prime }=I$
$\Rightarrow \left[\begin{array}{ccc}0& 2\alpha & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right]\times \left[\begin{array}{ccc}0& \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{ccc}4{\beta }^2+{\gamma }^2& 2{\beta }^2-{\gamma }^2& -2{\beta }^2+{\gamma }^2\\ 2{\beta }^2-{\gamma }^2& {\alpha }^2+{\beta }^2+{\gamma }^2& {\alpha }^2-{\beta }^2-{\gamma }^2\\ -2{\beta }^2+{\gamma }^2& {\alpha }^2-{\beta }^2-{\gamma }^2& {\alpha }^2+{\beta }^2+{\gamma }^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
Equating the corresponding elements, we have
$4{\beta }^2+{\gamma }^2=1...\left(1\right)$
$2{\beta }^2-{\gamma }^2=0...\left(2\right)$
${\alpha }^2+{\beta }^2+{\gamma }^2=1...\left(3\right)$
From $\left(1\right)$ and $\left(2\right)$, $6{\beta }^2=1\therefore {\beta }^2=\frac{1}{6}$
and ${\gamma }^2=\frac{1}{3}$
From
$\left(3\right){\alpha }^2=1-{\beta }^2-{\gamma }^2=1-\frac{1}{6}-\frac{1}{3}=\frac{1}{2}$
Hence, $\alpha =\pm \frac{1}{\sqrt{2}},\beta =\pm \frac{1}{\sqrt{6}}and\gamma =\pm \frac{1}{\sqrt{3}}$