Question

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

• A. $0$
• B. $1$
• C. ${\alpha }^2-{\beta }^2$
• D. ${\alpha }^2+{\beta }^2$

Answer 1

Answer: (A)

Let $\Delta =\left|\begin{array}{ccc}1& \cos (\alpha -\beta )& \cos \alpha \\ \cos (\alpha -\beta )& 1& \cos \beta \\ \cos \alpha & \cos \beta & 1\end{array}\right|$
Applying $R_2\to R_2-\cos (\alpha -\beta )R_1$ ,
$R_3\to R_3-\cos \alpha R_1$
$\left|\begin{array}{ccc}1& \cos (\alpha -\beta )& \cos \alpha \\ 0& 1-{\cos }^2(\alpha -\beta )& \cos \beta \cos \alpha \cos (\alpha -\beta )\\ 0& \cos \beta -\cos \alpha \cos (\alpha -\beta )& 1-{\cos }^2\alpha \end{array}\right|$
$=[1-{\cos }^2(\alpha -\beta )][1-{\cos }^2\alpha ]-[\cos \beta -\cos \alpha (\alpha -\beta ){]}^2$
$=1-{\cos }^2\alpha -{\cos }^2(\alpha -\beta )+{\cos }^2\alpha {\cos }^2(\alpha -\beta )$
$-{\cos }^2\beta -{\cos }^2\alpha {\cos }^2(\alpha -\beta )+2\cos \alpha \cos \beta$
$\cos (\alpha -\beta )=1-{\cos }^2\alpha -{\cos }^2\beta -{\cos }^2(\alpha -\beta )$
$+2\cos \alpha \cos \beta \cos (\alpha -\beta )$
$=1-{\cos }^2\alpha -{\cos }^2\beta -\cos (\alpha -\beta )[\cos (\alpha -\beta )$
$-2\cos \alpha \cos \beta ]$
$=1-{\cos }^2\alpha -{\cos }^2\beta -\cos (\alpha -\beta )$
$[\cos (\alpha -\beta )-\cos (\alpha +\beta )-\cos (\alpha -\beta )]$
$=1-{\cos }^2\alpha -{\cos }^2\beta -\cos (\alpha -\beta )$
$[-\cos (\alpha +\beta )]$
$=1-{\cos }^2\alpha -{\cos }^2\beta +\cos (\alpha -\beta )\cos (\alpha +\beta )$
$=1-{\cos }^2\alpha -{\cos }^2\beta +{\cos }^2\alpha -{\sin }^2\beta$
$=1-({\cos }^2\beta +{\sin }^2\beta )=1-1$
$=0$