Question

Using properties of determinants, prove that:
\(\begin{vmatrix} \alpha &\alpha^2  &\beta+\gamma \\   \beta&\beta^2  &\gamma+\alpha \\   \gamma&\gamma^2  &\alpha+\beta  \end{vmatrix}\)=(\(\beta-\gamma\))( \(\gamma-\alpha\))(\(\alpha-\beta\))(\(\alpha+\beta+\gamma\))

Answer 1

Δ=\(\begin{vmatrix} \alpha &\alpha^2  &\beta+\gamma \\   \beta&\beta^2  &\gamma+\alpha \\   \gamma&\gamma^2  &\alpha+\beta  \end{vmatrix}\)
Applying R₂\(\rightarrow\)R₂-R₁ and R₃\(\rightarrow\)R₃-R₁,we have
=\(\begin{vmatrix} \alpha &\alpha^2  &\beta+\gamma \\   \beta+\alpha&\beta^2-\alpha^2  &\alpha+\beta \\   \gamma-\alpha&\gamma^2-\alpha^2  &\alpha-\gamma  \end{vmatrix}\)

Applying R₃\(\rightarrow\)R₃-R₂, we have:
 Δ=(β-α)(γ-α)\(\begin{vmatrix} \alpha &\alpha^2  &\beta+\gamma \\   1&\beta+\alpha  &-1 \\   0&\gamma-\beta  &0  \end{vmatrix}\)

Expanding along R₃,we have:
Δ=(β-α)(γ-α)[-(γ-β)(-α-β-γ)]
=(β-α)(γ-α)(γ-β)(α+β+γ)
=(β-γ)( γ-α)(α-β)(α+β+γ)

Hence,the given result is proved.