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### Answers (1)

Answer: $\frac{1}{2}$

Hint: Here, we use basic concept of determinant of matrix.

Given: $\cos 2 \theta=0$

$\left[\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right]^{2}=?$

Solution:

$\cos 2 \theta=0$

\begin{aligned} &\cos 2 \theta=\cos 90^{\circ} \quad\left(\cos 90^{\circ}=0\right) \\ &2 \theta=90^{\circ} \\ &\theta=45^{\circ} \end{aligned}

Let’s put $\theta=45^{\circ}$  in matrix

$\left[\begin{array}{ccc} 0 & \cos 45^{\circ} & \sin 45^{\circ} \\ \cos 45^{\circ} & \sin 45^{\circ} & 0 \\ \sin 45^{\circ} & 0 & \cos 45^{\circ} \end{array}\right]^{2}$

$\left[\begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{array}\right]^{2} \quad\left[\sin 45^{\circ}=\frac{1}{\sqrt{2}}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]$

Let’s take $\frac{1}{\sqrt{2}}$ in common

$=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$

\begin{aligned} &=\left[\frac{1}{2 \sqrt{2}}[-1 \times(1-0)+1(0-1)]\right]^{2} \\ &=\frac{1}{8} \times[-1-1]^{2} \\ &=\frac{1}{8} \times(-2)^{2} \\ &=\frac{1}{2} \end{aligned}

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