Question

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

& A.p \\\ & B.q \\\ & C.{{p}^{2}}-2q \\\ & D.0 \\\ \end{aligned}$$

Answer 1

To solve this question, we will use three basic mathematical value which are given as below:
Sum of roots of a cubic (3 degree) equation is given by $\alpha +\beta +\gamma$ where equation is of the type $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$
We also know the formula, sum of roots $\Rightarrow \alpha +\beta +\gamma =\dfrac{-b}{a}=\dfrac{-\text{coefficient of }{{\text{x}}^{\text{2}}}}{\text{coefficient of }{{\text{x}}^{3}}}$
Using this, we will get the value of $\alpha +\beta +\gamma$ which can then be substituted in the determinant after applying row transformations.

Complete step-by-step answer:
We are given the equation as ${{x}^{3}}+px+q=0$.
Now, we know that sum of roots of a cubic (3 degree) equation can be written as $\alpha +\beta +\gamma$ where equation is of the type $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$. We also have a formula for finding the sum of roots. It is given as below,
$\Rightarrow \alpha +\beta +\gamma =\dfrac{-b}{a}=\dfrac{-\text{coefficient of }{{\text{x}}^{\text{2}}}}{\text{coefficient of }{{\text{x}}^{3}}}$
Here, in this given equation ${{x}^{3}}+px+q=0$ we do not have any term of ${{x}^{2}}$ so, the coefficient of ${{x}^{2}}=0$
Hence, we get the sum of roots of equation ${{x}^{3}}+px+q=0$ as

& \alpha +\beta +\gamma =0\text{ as coefficient of }{{\text{x}}^{\text{2}}}=0 \\\ & \alpha +\beta +\gamma =0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\\ \end{aligned}$$ Let us consider the determinant $$\left| \begin{matrix} \alpha & \beta & \gamma \\\ \beta & \gamma & \alpha \\\ \gamma & \alpha & \beta \\\ \end{matrix} \right|$$ Determinant property says that, sum of elements of row or column changes the value of determinant. Hence, adding all rows element of above determinant we get: $$\left| \begin{matrix} \alpha & \beta & \gamma \\\ \beta & \gamma & \alpha \\\ \gamma & \alpha & \beta \\\ \end{matrix} \right|=\left| \begin{matrix} \alpha +\beta +\gamma & \beta & \gamma \\\ \alpha +\beta +\gamma & \gamma & \alpha \\\ \alpha +\beta +\gamma & \alpha & \beta \\\ \end{matrix} \right|$$ Using equation (i) we get: $$\left| \begin{matrix} \alpha & \beta & \gamma \\\ \beta & \gamma & \alpha \\\ \gamma & \alpha & \beta \\\ \end{matrix} \right|=\left| \begin{matrix} 0 & \beta & \gamma \\\ 0 & \gamma & \alpha \\\ 0 & \alpha & \beta \\\ \end{matrix} \right|$$ Determinant property says that, if any one row or column is zero then the value of determinant is zero. $$\left| \begin{matrix} 0 & \beta & \gamma \\\ 0 & \gamma & \alpha \\\ 0 & \alpha & \beta \\\ \end{matrix} \right|=0$$ Therefore, the value of $$\left| \begin{matrix} \alpha & \beta & \gamma \\\ \beta & \gamma & \alpha \\\ \gamma & \alpha & \beta \\\ \end{matrix} \right|=0$$ Therefore, the answer is 0, **So, the correct answer is “Option D”.** **Note:** Another way to solve this question can be, opening the determinant $$\left| \begin{matrix} 0 & \beta & \gamma \\\ 0 & \gamma & \alpha \\\ 0 & \alpha & \beta \\\ \end{matrix} \right|$$ to get the required answer. Consider opening from first column, we get: $$\begin{aligned} & \left| \begin{matrix} 0 & \beta & \gamma \\\ 0 & \gamma & \alpha \\\ 0 & \alpha & \beta \\\ \end{matrix} \right|=0\left| \begin{matrix} \gamma & \alpha \\\ \alpha & \beta \\\ \end{matrix} \right|-0\left| \begin{matrix} \beta & \gamma \\\ \alpha & \beta \\\ \end{matrix} \right|+0\left| \begin{matrix} \beta & \gamma \\\ \gamma & \alpha \\\ \end{matrix} \right| \\\ & \Rightarrow \left| \begin{matrix} 0 & \beta & \gamma \\\ 0 & \gamma & \alpha \\\ 0 & \alpha & \beta \\\ \end{matrix} \right|=0-0-0 \\\ & \Rightarrow \left| \begin{matrix} 0 & \beta & \gamma \\\ 0 & \gamma & \alpha \\\ 0 & \alpha & \beta \\\ \end{matrix} \right|=0 \\\ \end{aligned}$$ Hence, the value of determinants: $$\left| \begin{matrix} \alpha & \beta & \gamma \\\ \beta & \gamma & \alpha \\\ \gamma & \alpha & \beta \\\ \end{matrix} \right|=0$$