Question

If $\alpha +\beta +\gamma =2\pi ,$ then the system of equations
$x+\left(\cos \gamma \right)y+\left(\cos \beta \right)z=0$
$\left(\cos \gamma \right)x+y+\left(\cos \alpha \right)z=0$
$\left(\cos \beta \right)x+\left(\cos \alpha \right)y+z=0$ has

• A. a unique solution
• B. no solution
• C. infinitely many solutions
• D. exactly two solutions

Answer 1

The correct option is C infinitely many solutions

$\Delta =\left|\begin{array}{ccc}1& \cos \gamma & \cos \beta \\ \cos \gamma & 1& \cos \alpha \\ \cos \beta & \cos \alpha & 1\end{array}\right|$
$=(1-{\cos }^2\alpha )-\cos \gamma (\cos \gamma -\cos \alpha \cos \beta )+\cos \beta (\cos \alpha \cos \gamma -\cos \beta )$
$=1-({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma )+2\cos \alpha \cos \beta \cos \gamma$

As $\alpha +\beta +\gamma =2\pi$
$\Rightarrow \alpha +\beta =2\pi -\gamma$
$\Rightarrow \cos (\alpha +\beta )=\cos (2\pi -\gamma )$
$\Rightarrow \cos \alpha \cos \beta -\sin \alpha \sin \beta =\cos \gamma$
$\Rightarrow \cos \alpha \cos \beta -\cos \gamma =\sin \alpha \sin \beta$
Squaring both the sides
$\Rightarrow (\cos \alpha \cos \beta -\cos \gamma {)}^2=(\sin \alpha \sin \beta {)}^2$
$\Rightarrow {\cos }^2\alpha {\cos }^2\beta +{\cos }^2\gamma -2\cos \alpha \cos \beta \cos \gamma =(1-{\cos }^2\alpha )(1-{\cos }^2\beta )$
$\Rightarrow {\cos }^2\alpha {\cos }^2\beta +{\cos }^2\gamma -2\cos \alpha \cos \beta \cos \gamma =1-{\cos }^2\alpha -{\cos }^2\beta +{\cos }^2\alpha {\cos }^2\beta$
$\Rightarrow 1-({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma )+2\cos \alpha \cos \beta \cos \gamma =0$

$\Delta =0$
$\therefore$ System has infinitely many solutions.