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Matrices & Determinants: Roots System Possesses Non Zero Solutions Equals

JEE Maths question with a full step-by-step solution.

Question
If α,β,γ\alpha, \beta, \gamma are the roots of x3+ax2+bx+c=0x^3+ax^2+bx+c=0 and if the system αu+βv+γw=0\alpha u+\beta v+\gamma w = 0, βu+γv+αw=0\beta u+\gamma v+\alpha w = 0, γu+αv+βw=0\gamma u+\alpha v+\beta w = 0 possesses non-zero solutions, and a3−6c=0a^3-6c=0, then 2c2c equals:
Aababcorrect
B2ab2ab
C3ab3ab
DNone of these
Solution
Step 1: Condition for a non-trivial solution The system has a non-trivial solution if and only if:
∣αβγβγαγαβ∣=0\begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} = 0
 Step 2: Evaluate using the circulant determinant identity
det⁡=α3+β3+γ3−3αβγ=(α+β+γ)[(α+β+γ)2−3(αβ+βγ+γα)]=0\det = \alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma = (\alpha+\beta+\gamma)\bigl[(\alpha+\beta+\gamma)^2-3(\alpha\beta+\beta\gamma+\gamma\alpha)\bigr] = 0
 Substituting Vieta's relations α+β+γ=−a\alpha+\beta+\gamma=-a and αβ+βγ+γα=b\alpha\beta+\beta\gamma+\gamma\alpha=b:
(−a)(a2−3b)=0  ⟹  a3−3ab=0(-a)(a^2-3b) = 0 \implies a^3-3ab = 0
 Step 3: Combine with the given condition From a3−3ab=0a^3 - 3ab = 0 and a3−6c=0a^3 - 6c = 0:
−3ab+6c=0  ⟹  2c=ab-3ab + 6c = 0 \implies 2c = ab
 Answer: (1)
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