The value of the determinant | cos α -sin α 1 sin α cos α 1 cos (α+β) -sin (α+β) 1 | is

Correct answer: (A) independent of α. Given, | cos α -sin α 1 sin α cos α 1 cos (α+β) -sin (α+β) 1 | [ . Applying .R_3 arrow R_3-R_1(cos β)+R_2(sin β) ] = | cos α -sin α 1 sin α cos α 1 0 0 1+sin β-cos β | =(1+sin β-cos β) (cos^2 α+sin^2 α ) =1+sin β-cos β , which is independent of α .

Q59 / 120 · VITEEE 2012 · Mathematics

The value of the determinant $\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos (\alpha+\beta) & -\sin (\alpha+\beta) & 1\end{array}\right|$ is

Correct answer
A
independent of $\alpha$
B
independent of $\beta$
C
independent of $\alpha$ and $\beta$
D
None of the above

Solution

Given, $\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos (\alpha+\beta) & -\sin (\alpha+\beta) & 1\end{array}\right|$ $\left[\right.$ Applying $\left.\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}(\cos \beta)+\mathrm{R}_{2}(\sin \beta)\right]$ $=\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ 0 & 0 & 1+\sin \beta-\cos \beta\end{array}\right|$ $=(1+\sin \beta-\cos \beta)\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)$ $=1+\sin \beta-\cos \beta$, which is independent of $\alpha .$

Hence, the correct answer is option: 'A'.

The value of the determinant $\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos (\alpha+\beta) & -\sin (\alpha+\beta) & 1\end{array}\right|$ is

Solution

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