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Question: The values of the determinant \(\left| \begin{matrix} 1 & \cos(\alpha - \beta) & \cos\alpha \\ \cos...

The values of the determinant

∣1cos⁡(α−β)cos⁡αcos⁡(α−β)1cos⁡βcos⁡αcos⁡β1∣\left| \begin{matrix} 1 & \cos(\alpha - \beta) & \cos\alpha \\ \cos(\alpha - \beta) & 1 & \cos\beta \\ \cos\alpha & \cos\beta & 1 \end{matrix} \right|is.

A

α2+β2\alpha^{2} + \beta^{2}

B

α2−β2\alpha^{2} - \beta^{2}

C

1

D

0

Answer

0

Explanation

Solution

On solving the determinant,

1(1−cos⁡2β)−cos⁡(α−β)[cos⁡(α−β)−cos⁡αcos⁡β]1(1 - \cos^{2}\beta) - \cos(\alpha - \beta)\lbrack\cos(\alpha - \beta) - \cos\alpha\cos\beta\rbrack

+cos⁡α[cos⁡βcos⁡(α−β)−cos⁡α]+ \cos\alpha\lbrack\cos\beta\cos(\alpha - \beta) - \cos\alpha\rbrack

=1−cos⁡2β−cos⁡2α−cos⁡2(α−β)+2cos⁡αcos⁡βcos⁡(α−β)= 1 - \cos^{2}\beta - \cos^{2}\alpha - \cos^{2}(\alpha - \beta) + 2\cos\alpha\cos\beta\cos(\alpha - \beta)

= 1−cos⁡2β−cos⁡2α+cos⁡(α−β)1 - \cos^{2}\beta - \cos^{2}\alpha + \cos(\alpha - \beta) (2cos⁡αcos⁡β−cos⁡(α−β))(2\cos\alpha\cos\beta - \cos(\alpha - \beta))

= 1−cos⁡2β−cos⁡2α+cos⁡(α−β)1 - \cos^{2}\beta - \cos^{2}\alpha + \cos(\alpha - \beta)

[cos⁡(α+β)+cos⁡(α−β)−cos⁡(α−β)]\lbrack\cos(\alpha + \beta) + \cos(\alpha - \beta) - \cos(\alpha - \beta)\rbrack

= 1−cos⁡2β−cos⁡2α+cos⁡(α−β)cos⁡(α+β)1 - \cos^{2}\beta - \cos^{2}\alpha + \cos(\alpha - \beta)\cos(\alpha + \beta)

= 1−cos⁡2β−cos⁡2α+cos⁡2αcos⁡2β−sin⁡2αsin⁡2β1 - \cos^{2}\beta - \cos^{2}\alpha + \cos^{2}\alpha\cos^{2}\beta - \sin^{2}\alpha\sin^{2}\beta

= 1−cos⁡2β−cos⁡2α(1−cos⁡2β)−sin⁡2αsin⁡2β1 - \cos^{2}\beta - \cos^{2}\alpha(1 - \cos^{2}\beta) - \sin^{2}\alpha\sin^{2}\beta

= 1−cos⁡2β−cos⁡2αsin⁡2β−sin⁡2αsin⁡2β1 - \cos^{2}\beta - \cos^{2}\alpha\sin^{2}\beta - \sin^{2}\alpha\sin^{2}\beta

= 1−cos⁡2β−sin⁡2β1 - \cos^{2}\beta - \sin^{2}\beta = 0.