Matrices and Determinants 2 Question 3
3. If $\Delta_{1}=\left|\begin{matrix}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{matrix}\right|$ and $\Delta_{2}=\left|\begin{matrix}x & \sin 2 \theta & \cos 2 \theta \\ -\sin 2 \theta & -x & 1 \\ \cos 2 \theta & 1 & x\end{matrix}\right|, x \neq 0$, then for all $\theta \in 0, \frac{\pi}{2}$
(a) $\Delta_{1}+\Delta_{2}=-2\left(x^{3}+x-1\right)$
(b) $\Delta_{1}-\Delta_{2}=-2 x^{3}$
(c) $\Delta_{1}+\Delta_{2}=-2 x^{3}$
(d) $\Delta_{1}-\Delta_{2}=x(\cos 2 \theta-\cos 4 \theta)$
(2019 Main, 10 April I)
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Answer:
Correct Answer: 3. (c)
Solution:
- Given determinants are
$ \Delta_{1}=\left|\begin{matrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{matrix}\right| $
$ \begin{aligned} & =-x^{3}+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^{2} \theta-x+x \sin ^{2} \theta \\ & =-x^{3} \\ & \text { and } \Delta_{2}=\left|\begin{matrix} x & \sin 2 \theta & \cos 2 \theta \\ -\sin 2 \theta & -x & 1 \\ \cos 2 \theta & 1 & x \end{matrix}\right|, x \neq 0 \\ & =-x^{3}\left(\text { similarly as } \Delta_{1}\right) \end{aligned} $
So, according to options, we get $\Delta_{1}+\Delta_{2}=-2 x^{3}$
