Question

The value of

$\left|\begin{array}{ccc}2& \alpha +\beta +\gamma +\delta & \alpha \beta +\gamma \delta \\ \alpha +\beta +\gamma +\delta & 2(\alpha +\beta )(\gamma +\delta )& \alpha \beta (\gamma +\delta )+\gamma \delta (\alpha +\beta )\\ \alpha \beta +\gamma \delta & \alpha \beta (\gamma +\delta )+\gamma \delta (\alpha +\beta )& 2\alpha \beta \gamma \delta \end{array}\right|$

Answer 1

$L.H.S=\left|\begin{array}{ccc}2& \alpha +\beta +\gamma +\delta & \alpha \beta +\gamma \delta \\ \alpha +\beta +\gamma +\delta & 2(\alpha +\beta )(\gamma +\delta )& \alpha \beta (\gamma +\delta )+\gamma \delta (\alpha +\beta )\\ \alpha \beta +\gamma \delta & \alpha \beta (\gamma +\delta )+\gamma \delta (\alpha +\beta )& 2\alpha \beta \gamma \delta \end{array}\right|=\left|\begin{array}{ccc}1& 1& 0\\ \alpha +\beta & \gamma +\delta & 0\\ \alpha \beta & \gamma \delta & 0\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 0\\ \gamma +\delta & \alpha +\beta & 0\\ \gamma \delta & \alpha \beta & 0\end{array}\right|\left(RowbyRow\right)=0=R.H.S$