Question

If the matrix $â\begin{array}{ccc}0& 2β& γ\\ Î\pm & β& −γ\\ Î\pm & −β& γ\end{array}â$ is orthogonal, then

• A. $Î\pm =Â\pm \frac{1}{\sqrt{2}}$
• B. $β=Â\pm \frac{1}{\sqrt{6}}$
• C. $γ=Â\pm \frac{1}{\sqrt{3}}$
• D. All of these

Answer 1

Answer: (D)

Let $A=â\begin{array}{ccc}0& 2β& γ\\ Î\pm & β& −γ\\ Î\pm & =β& γ\end{array}â$ Since, matrix $A$ is orthogonal $∴A{A}^{′}=1⇒$

\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}

$â\begin{array}{ccc}0& 2β& γ\\ Î\pm & β& −γ\\ Î\pm & −β& γ\end{array}ââ\begin{array}{ccc}0& Î\pm & Î\pm \\ 2β& β& −β\\ γ& −γ& γ\end{array}â=â\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}â$ $⇒â\begin{array}{ccc}4{β}^2+{γ}^2& 2{β}^2−{γ}^2& −2{β}^2+{γ}^2\\ 2{β}^2−{γ}^2& {Î\pm }^2+{β}^2+{γ}^2& {Î\pm }^2−{β}^2−{γ}^2\\ −2{β}^2+{γ}^2& {Î\pm }^2−{β}^2−{γ}^2& {Î\pm }^2+{β}^2+{γ}^2\end{array}â$ $=â\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}â$ Equating the corresponding elements of above matrices, we get $4{β}^2+{γ}^2=1?$ (i) $2{β}^2−{γ}^2=0…$ (ii) ${Î\pm }^2+{β}^2+{γ}^2=1$ ? (iii) Adding Eqs. (i) and (ii), we get $6{β}^2=1⇒β=Â\pm \frac{1}{\sqrt{6}}$ From Eq. (ii), $⇒{γ}^2=2{β}^2⇒{γ}^2=\frac{2}{6}=\frac{1}{3}⇒γ=Â\pm \frac{1}{\sqrt{3}}$
From Eq. (iii), ${Î\pm }^2=1−{β}^2−{γ}^2⇒$
${Î\pm }^2=1−\frac{1}{6}−\frac{1}{3}=\frac{1}{2}⇒Î\pm =Â\pm \frac{1}{\sqrt{2}}$