Question

If the matrix $\left[\begin{array}{ccc}0& 2\beta & {\rm Y}\\ \alpha & \beta & -{\rm Y}\\ \alpha & -\beta & {\rm Y}\end{array}\right]$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

The correct option is D all of these

$\left[\begin{array}{ccc}0& 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right]$

for orthogonal matrix we have

$A.A^T=I$

$\left[\begin{array}{ccc}0& 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right]\left[\begin{array}{ccc}0& \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\left[\begin{array}{ccc}0+4{\beta }^2+{\gamma }^2& 0+2{\beta }^2-{\gamma }^2& 0-2{\beta }^2+{\gamma }^2\\ 0+2{\beta }^2-{\gamma }^2& {\alpha }^2+{\beta }^2+{\gamma }^2& {\alpha }^2-{\beta }^2-{\gamma }^2\\ 0-2{\beta }^2+{\gamma }^2& {\alpha }^2-{\beta }^2-{\gamma }^2& {\alpha }^2+{\beta }^2+{\gamma }^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$4{\beta }^2+{\gamma }^2=1,2{\beta }^2-{\gamma }^2=0$,

$4\left(\frac{{\gamma }^2}{2}\right)+{\gamma }^2=1$ ${\beta }^2=\frac{r^2}{2}$

$r^2\left[3\right]=1$

$r=\pm \frac{1}{\sqrt{3}}$

$2{\beta }^2-{\gamma }^2=0,{\alpha }^2+{\beta }^2+{\gamma }^2+{\gamma }^2=1,{\alpha }^2-{\beta }^2-{\gamma }^2=0$

${\beta }^2=\frac{{\gamma }^2}{2},{\alpha }^2+\frac{{\gamma }^2}{2}+\frac{{\gamma }^2}{1}=1$

${\alpha }^2+\frac{3{\gamma }^2}{2}=1$

${\beta }^2=\frac{1}{6}{\alpha }^2+\frac{3}{2}\times \frac{1}{3}=1$

$\beta =\pm \frac{1}{\sqrt{6}}$ $\alpha =\pm \frac{1}{\sqrt{2}}$