Mathematics - Determinant Question with Solution | TestHub

Given, $\alpha +\beta +\gamma =2\pi$

Now, $\mathrm{\Delta }=\left|\begin{array}{ccc}1& \cos \gamma & \cos \beta \\ \cos \gamma & 1& \cos \alpha \\ \cos \beta & \cos \alpha & 1\end{array}\right|$

By expanding along the first column, we get

$∆=1\left(1-{\cos }^2\alpha \right)-\cos \gamma \left(\cos \gamma -\cos \alpha .\cos \beta \right)+\cos \beta \left(\cos \alpha .\cos \gamma -\cos \beta \right)$

$∆=1-{\cos }^2\alpha -{\cos }^2\gamma +\cos \gamma .\cos \alpha .\cos \beta +\cos \beta .\cos \alpha .\cos \gamma -{\cos }^2\beta$

$=1-{\cos }^2\alpha -{\cos }^2\beta -{\cos }^2\gamma +2\cos \alpha .\cos \beta .\cos \gamma$

$={\sin }^2\alpha -{\cos }^2\beta -\cos \gamma (\cos \gamma -2\cos \alpha .\cos \beta )$

$=-\cos (\alpha +\beta ).\cos (\alpha -\beta )-\cos \gamma \left[\cos \left\{2\pi -(\alpha +\beta )\right\}-2\cos \alpha .\cos \beta \right]$

$=-\cos (2\pi -\gamma )\cos (\alpha -\beta )-\cos \gamma \left[\cos (\alpha +\beta )-2\cos \alpha .\cos \beta \right]$

$=-\cos (2\pi -\gamma ).\cos (\alpha -\beta )-\cos \gamma \left[\cos \alpha .\cos \beta -\sin \alpha .\sin \beta -2\cos \alpha .\cos \beta \right]$

$=-\cos \gamma .\cos (\alpha -\beta )+\cos \gamma .\cos (\alpha -\beta )$

$=0$

So, $∆=0$

Hence, the system of equations has infinitely many solutions.