Question

If Δ = `|(1, x, x^2),(1, y, y^2),(1, z, z^2)|`, Δ₁ = `|(1, 1, 1),(yz, zx, xy),(x, y, z)|`, then prove that ∆ + ∆₁ = 0.

Answer 1

We have Δ₁ = `|(1, 1, 1),(yz, zx, xy),(x, y, z)|`

Interchanging rows and columns, we get

Δ₁ = `|(1, yz, x),(1, z, y),(1, xy, z)|`

= `1/(xyz) |(x, xyz, x^2),(y, xyz, y^2),(z, xyz, z^2)|`

= `(xyz)/(xyz) |(x, 1, x^2),(y, 1, y^2),(z, 1, z^2)|`

Interchanging C₁ and C₂

= `(-1)|(1, x, x^2),(1, y, y^2),(1, z, z^2)|`

= – Δ

⇒ Δ₁ + Δ = 0