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Mathematics Question on Determinants

The value of the determinant ∣cos⁡α−sin⁡α1 sin⁡αcos⁡α1 cos⁡(α+β)−sin⁡α+β1∣\begin{vmatrix}\cos\alpha&-\sin\alpha&1\\\ \sin \alpha&\cos\alpha&1\\\ \cos\left(\alpha+\beta\right)&-\sin\alpha+\beta&1\end{vmatrix} is

A

independent of α\alpha

B

independent of β\beta

C

independent of α\alpha and β\beta

D

None of the above

Answer

independent of α\alpha

Explanation

Solution

Given, ∣cos⁡α−sin⁡α1 sin⁡αcos⁡α1 cos⁡(α+β)−sin⁡α+β1∣\begin{vmatrix}\cos\alpha&-\sin\alpha&1\\\ \sin \alpha&\cos\alpha&1\\\ \cos\left(\alpha+\beta\right)&-\sin\alpha+\beta&1\end{vmatrix}
[Applying R3→R3−R1(cos⁡β)+R2(sin⁡β)]R_{3} \to R_{3} - R_{1}\left(\cos\beta\right) +R_{2}\left(\sin\beta\right) ]
=(cos⁡α−sin⁡α1 sin⁡αcos⁡α1 001+sin⁡β−cos⁡β)= \begin{pmatrix}\cos\alpha&-\sin\alpha&1\\\ \sin\alpha&\cos\alpha&1\\\ 0&0&1+\sin\beta-\cos\beta\end{pmatrix}
=(1+sin⁡β−cos⁡β)(cos⁡2α+sin⁡2α)= \left(1+\sin\beta -\cos\beta\right)\left(\cos^{2} \alpha + \sin^{2} \alpha\right)
=1+sin⁡β−cos⁡β= 1+\sin\beta-\cos\beta which is independent of α\alpha