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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS1 Mark
Question
Fill in the blanks:
If $\cos2\theta=0,$ then $\begin{vmatrix}0&\cos\theta&\sin\theta\\\cos\theta&\sin\theta&0\\\sin\theta&0&\cos\theta\end{vmatrix}^2=$ _________.
✓

Answer

If $\cos2\theta=0,$ then $\begin{vmatrix}0&\cos\theta&\sin\theta\\\cos\theta&\sin\theta&0\\\sin\theta&0&\cos\theta\end{vmatrix}^2=\frac{1}{2}.$
Solution:
Since, $\cos2\theta=0$
$\Rightarrow\ \cos2\theta=\cos\frac{\pi}{2}$ $\Rightarrow\ 2\theta=\frac{\pi}{2}$
$\Rightarrow\ \theta=\frac{\pi}{4}$
$\therefore\ \sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}$ and $\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$
$\therefore\ \begin{vmatrix}0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\end{vmatrix}$
Expanding along $R_1$,
$=\bigg[-\frac{1}{\sqrt{2}}\Big(\frac{1}{2}\Big)+\frac{1}{\sqrt{2}}\Big(-\frac{1}{2}\Big)\bigg]^2$ $=\bigg[\frac{-2}{2\sqrt{2}}\bigg]^2=\bigg(\frac{-1}{\sqrt{2}}\bigg)^2=\frac{1}{2}$

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