Matrices and Determinants 2 Question 3
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3. If $\Delta _1=\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \ -\sin \theta & -x & 1 \ \cos \theta & 1 & x\end{array}\right|$ and $\Delta _2=\left|\begin{array}{ccc}x & \sin 2 \theta & \cos 2 \theta \ -\sin 2 \theta & -x & 1 \ \cos 2 \theta & 1 & x\end{array}\right|, x \neq 0$, then for all $\theta \in 0, \frac{\pi}{2}$
======= ####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 $\theta \in 0, \frac{\pi}{2}$
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(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 $\Delta _1+\Delta _2=-2 x^{3}$
