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Mathematics Question on Determinants

If α,β,γ\alpha , \beta , \gamma are the roots of the equation x3+px+q=0x^3 + px + q = 0, then the value of the determinant ∣αβγ βγα γαβ∣=\begin{vmatrix}\alpha&\beta&\gamma\\\ \beta&\gamma&\alpha\\\ \gamma&\alpha&\beta\end{vmatrix} =

A

qq

B

0

C

pp

D

p2−2pp^2 - 2p

Answer

0

Explanation

Solution

We have α,β,γ\alpha , \beta , \gamma are the? roots of equation x3+px+q=0     x^3 + px + q = 0 \ \ \ \ \ ...(i)
Sum of roots = α+β+γ=0 αβ+βγ+γα=p   \alpha + \beta + \gamma = 0 \: \alpha \beta + \beta \gamma + \gamma \alpha = p \ \ \ ...(ii)
Product of roots = αβγ=−q\alpha \beta \gamma = - q
Applying C1→C1+C2+C3,C_1 \to C_1 + C_2 + C_3, we get
=∣α+β+γβγ α+β+γγα α+β+γαβ∣= \begin{vmatrix}\alpha+\beta+\gamma&\beta&\gamma\\\ \alpha +\beta +\gamma &\gamma&\alpha\\\ \alpha +\beta +\gamma &\alpha&\beta\end{vmatrix}
=(α+β+γ)∣1βγ 1γα 1αβ∣= \left(\alpha+\beta+\gamma\right) \begin{vmatrix}1&\beta&\gamma\\\ 1&\gamma&\alpha\\\ 1&\alpha&\beta\end{vmatrix}
Applying R2→R2−R1,R3→R3−R1R_2 \to R2 - R_1, R_3 \to R_3 - R_1, we get
=(α+β+γ)∣1βγ 0γ−βα−γ 0α−ββ−γ∣= \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 C1C_1, we get
    =(α+β+γ)(−(β−γ)2−(α−β)(α−γ))\ \ \ \ = (\alpha + \beta +\gamma)(- (\beta - \gamma)^2 - (\alpha - \beta )(\alpha -\gamma))
    =0\ \ \ \ = 0