ISI2017-MMA-19

Question

If $\alpha, \beta$ and $\gamma$ are the roots of $x^3 - px +q = 0$, then the value of the determinant

$$\begin{vmatrix}\alpha & \beta & \gamma\\\beta & \gamma & \alpha\\\gamma & \alpha & \beta\end{vmatrix}$$

is

• A. $p$
• B. $p^2$
• C. $0$
• D. $p^2+6q$

Answer 1

$\alpha ,\beta ,\gamma$ are roots of equation.

$\alpha +\beta +\gamma=0$

$\alpha \beta + \beta \gamma+ \gamma\alpha =-p$

$\alpha \cdot\beta \cdot\gamma=q$

The determinant of the matrix after simplification is

$3(\alpha \cdot\beta \cdot\gamma)-(\alpha^{3}+\beta^{3}+\gamma^{3})$

$3q-(\alpha^{3}+\beta^{3}+\gamma^{3}) $ ------------------$>(1)$

$\alpha^{3}+\beta^{3}+\gamma^{3}=(\alpha +\beta +\gamma)(\alpha^{2}+\beta^{2}+\gamma^{2}-\alpha \beta - \beta \gamma- \gamma\alpha )+3(\alpha \cdot\beta \cdot\gamma)=3q$

substitute the value of $\alpha^{3}+\beta^{3}+\gamma^{3}$ in equation  $1$ and  we get

$3q-3q=0$

So$,(C)$ is the correct choice.

Comments

No need to expand determinant.

$\begin{vmatrix} \alpha &\beta &\gamma \\ \beta &\gamma &\alpha \\ \gamma &\alpha &\beta \end{vmatrix}$

Now, $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$

$\begin{vmatrix} \alpha+\beta+\gamma &\beta &\gamma \\ \alpha+\beta+\gamma &\gamma &\alpha \\ \alpha+\beta+\gamma &\alpha &\beta \end{vmatrix}$

$(\alpha+\beta+\gamma)$ $\begin {vmatrix} 1 &\beta &\gamma \\ 1 &\gamma &\alpha \\ 1 &\alpha &\beta \end{vmatrix}$

Since, $(\alpha+\beta+\gamma)$ $=0$ here

So, $0 \times $$\begin {vmatrix} 1 &\beta &\gamma \\ 1 &\gamma &\alpha \\ 1 &\alpha &\beta \end{vmatrix}$ $= 0$

Answer 2

From equation alpha*beta*gamma=q

                         alpha+beta+gamma=0

 

MOD of the determinant=alpha^3+beta^3+gamma^3-(3*alpha*beta*gamma)

                                       =(alpha+beta+gamma)*(alpha^2+beta^2+gamma^2-alpha*beta-gamma*alpha-beta*gamma)

                                      =0

Answer (C)

Comments

use equation tool for more clearance

---

Please elaborate your answer