Question

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

• 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
$= \begin{vmatrix}\alpha+\beta+\gamma&\beta&\gamma\\ \alpha +\beta +\gamma &\gamma&\alpha\\ \alpha +\beta +\gamma &\alpha&\beta\end{vmatrix} $
$= \left(\alpha+\beta+\gamma\right) \begin{vmatrix}1&\beta&\gamma\\ 1&\gamma&\alpha\\ 1&\alpha&\beta\end{vmatrix}$
Applying $R_2 \to R2 - R_1, R_3 \to R_3 - R_1$, we get
$= \left(\alpha +\beta +\gamma \right) \begin{vmatrix}1&\beta &\gamma \\ 0&\gamma-\beta &\alpha -\gamma\\ 0&\alpha-\beta &\beta-\gamma \end{vmatrix}$
Now, expanding along $C_1$, we get
$ \ \ \ \ = (\alpha + \beta +\gamma)(- (\beta - \gamma)^2 - (\alpha - \beta )(\alpha -\gamma))$
$ \ \ \ \ = 0$