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
Correct Answer: D
Solution :
Let\[A=\left[ \begin{matrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & =\beta & \gamma \\ \end{matrix} \right]\] Since, matrix \[A\] is orthogonal \[\therefore \] \[AA'=1\] \[\Rightarrow \]\[\left[ \begin{matrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \\ \end{matrix} \right]\left[ \begin{matrix} 0 & \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \]\[\left[ \begin{matrix} 4{{\beta }^{2}}+{{\gamma }^{2}} & 2{{\beta }^{2}}-{{\gamma }^{2}} & -2{{\beta }^{2}}+{{\gamma }^{2}} \\ 2{{\beta }^{2}}-{{\gamma }^{2}} & {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} & {{\alpha }^{2}}-{{\beta }^{2}}-{{\gamma }^{2}} \\ -2{{\beta }^{2}}+{{\gamma }^{2}} & {{\alpha }^{2}}-{{\beta }^{2}}-{{\gamma }^{2}} & {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] Equating the corresponding elements of above matrices, we get \[4{{\beta }^{2}}+{{\gamma }^{2}}=1\] ? (i) \[2{{\beta }^{2}}-{{\gamma }^{2}}=0\] ... (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), \[{{\alpha }^{2}}=1-{{\beta }^{2}}-{{\gamma }^{2}}\] \[\Rightarrow \] \[{{\alpha }^{2}}=1-\frac{1}{6}-\frac{1}{3}=\frac{1}{2}\] \[\Rightarrow \] \[\alpha =\pm \frac{1}{\sqrt{2}}\]
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