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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.
Solution
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
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