Question

The value of the determinant $ \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. $ 0 $
• B. $ 1 $
• C. $\alpha^{2}-\beta^{2}$
• D. $\alpha^{2}+\beta^{2}$

Answer 1

Answer: (A)

Let $ \Delta = \begin{vmatrix} 1 & \cos (\alpha -\beta ) & \cos \alpha \\ \cos (\alpha -\beta ) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \\ \end{vmatrix} $
Applying $ R_{2}\to R_{2}-\cos (\alpha -\beta )R_{1} $ ,
$ R_{3}\to R_{3}-\cos \alpha R_{1} $
$\begin{vmatrix}1 & \cos (\alpha-\beta) & \cos \alpha \\ 0 & 1-\cos ^{2}(\alpha-\beta) & \cos \beta \cos \alpha \cos (\alpha-\beta) \\ 0 & \cos \beta-\cos \alpha \cos (\alpha-\beta) & 1-\cos ^{2} \alpha\end{vmatrix}$
$ =[1-\cos^{2}(\alpha -\beta )][1-\cos^{2}\alpha ] -[\cos \beta -\cos \alpha (\alpha -\beta )]^{2} $
$=1-\cos^{2}\alpha -\cos ^{2}(\alpha -\beta )+\cos ^{2}\alpha \cos ^{2}(\alpha -\beta ) $
$ -\cos ^{2}\beta -\cos ^{2}\alpha \cos ^{2}(\alpha -\beta )+2\cos \alpha \cos \beta $
$ \cos (\alpha -\beta ) =1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta -{{\cos }^{2}}(\alpha -\beta ) $
$ +2\cos \alpha \cos \beta \cos (\alpha -\beta ) $
$ =1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta -\cos (\alpha -\beta )[\cos (\alpha -\beta ) $
$ -2\cos \alpha \cos \beta ] $
$ =1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta -\cos (\alpha -\beta ) $
$ [\cos (\alpha -\beta )-\cos (\alpha +\beta )-\cos (\alpha -\beta )] $
$ =1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta -\cos (\alpha -\beta ) $
$ [-\cos (\alpha +\beta )] $
$ =1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta +\cos (\alpha -\beta )\cos (\alpha +\beta ) $
$ =1-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta +{{\cos }^{2}}\alpha -{{\sin }^{2}}\beta $
$ =1-({{\cos }^{2}}\beta +{{\sin }^{2}}\beta )=1-1 $
$ =0 $