Using properties of determinants, prove the following:
$|\begin{matrix} α & β & γ \\ α^{2} & β^{2} & γ^{2} \\ β + γ & γ + α & α + β \end{matrix}| = (α - β) (β - γ) (γ - α) (α + β + γ)$

Solution

$∆ = |\begin{matrix} α & β & γ \\ α^{2} & β^{2} & γ^{2} \\ β + γ & γ + α & α + β \end{matrix}| Applying R_{3} \to R_{3} + R_{1} ∆ = |\begin{matrix} α & β & γ \\ α^{2} & β^{2} & γ^{2} \\ α + β + γ & α + β + γ & α + β + γ \end{matrix}| = α + β + γ |\begin{matrix} α & β & γ \\ α^{2} & β^{2} & γ^{2} \\ 1 & 1 & 1 \end{matrix}| Applying C_{1} \to C_{1} - C_{2} and C_{2} \to C_{2} - C_{3}$

$∆ = α + β + γ = |\begin{matrix} α - β & β - γ & γ \\ α^{2} - β^{2} & β^{2} - γ^{2} & γ^{2} \\ 0 & 0 & 0 \end{matrix}| = (α + β + γ) (α - β) (β - γ) |\begin{matrix} 1 & 1 & 1 \\ α + β & β + γ & γ \\ 0 & 0 & 0 \end{matrix}| = (α + β + γ) (α - β) (β - γ) [1 (β + γ) - 1 (α + β)] = (α - β) (β - γ) (α + β + γ) (β + γ - α - β) = (α - β) (β - γ) (γ - α) (α + β + γ)$

Hence proved.

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Determinants

Using properties of determinants, prove the following:
$|\begin{matrix} α & β & γ \\ α^{2} & β^{2} & γ^{2} \\ β + γ & γ + α & α + β \end{matrix}| = (α - β) (β - γ) (γ - α) (α + β + γ)$

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Determinants

Using properties of determinants, prove the following:αβγα2β2γ2β + γ γ + α α + β = α - β β - γ γ - α α + β + γ

Some More Questions From Determinants Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Using properties of determinants, prove the following:
$|\begin{matrix} α & β & γ \\ α^{2} & β^{2} & γ^{2} \\ β + γ & γ + α & α + β \end{matrix}| = (α - β) (β - γ) (γ - α) (α + β + γ)$