Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+px+q=0$, then the value of the determinant $\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|=$

• A. $q$
• B. 0
• C. $p$
• D. $p^2-2p$

Answer 1

Answer: (B)

We have $\alpha ,\beta ,\gamma$ are the· roots of equation $x^3+px+q=0$...(i)
Sum of roots = $\alpha +\beta +\gamma =0\alpha \beta +\beta \gamma +\gamma \alpha =p$ ...(ii)
Product of roots = $\alpha \beta \gamma =-q$
Applying $C_1\to C_1+C_2+C_3,$ we get
$=\left|\begin{array}{ccc}\alpha +\beta +\gamma & \beta & \gamma \\ \alpha +\beta +\gamma & \gamma & \alpha \\ \alpha +\beta +\gamma & \alpha & \beta \end{array}\right|$
$=\left(\alpha +\beta +\gamma \right)\left|\begin{array}{ccc}1& \beta & \gamma \\ 1& \gamma & \alpha \\ 1& \alpha & \beta \end{array}\right|$
Applying $R_2\to R2-R_1,R_3\to R_3-R_1$, we get
$=\left(\alpha +\beta +\gamma \right)\left|\begin{array}{ccc}1& \beta & \gamma \\ 0& \gamma -\beta & \alpha -\gamma \\ 0& \alpha -\beta & \beta -\gamma \end{array}\right|$
Now, expanding along $C_1$, we get
$=(\alpha +\beta +\gamma )(-(\beta -\gamma {)}^2-(\alpha -\beta )(\alpha -\gamma ))$
$=0$