Question

Using properties of determinants,solve following equation
$\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\\ \beta +\gamma & \gamma +\alpha & \alpha +\beta \end{array}\right|=x\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\\ 1& 1& 1\end{array}\right|$

• A. $0$
• B. $(\alpha -\beta -\gamma )$
• C. $(\alpha +\beta +\gamma )$
• D. None of these

Answer 1

Answer: (C)

$(\alpha +\beta +\gamma )$

$\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\\ \beta +\gamma & \gamma +\alpha & \alpha +\beta \end{array}\right|=x\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\\ 1& 1& 1\end{array}\right|$

$R_3\to R_3+R_1$

$\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\\ \alpha +\beta +\gamma & \alpha +\beta +\gamma & \alpha +\beta +\gamma \end{array}\right|=x\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\\ 1& 1& 1\end{array}\right|$

Taking out common $\left(\alpha +\beta +\gamma \right)$ from $R_3$

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

$x=\left(\alpha +\beta +\gamma \right)$