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Question

Let ϕ1(x)=x+a1,ϕ2(x)=x2+b1x+b2,x1=2,x2=3,x3=5
Δ=∣ ∣ ∣111ϕ1(x1)ϕ1(x2)ϕ1(x3)ϕ2(x1)ϕ2(x2)ϕ2(x3)∣ ∣ ∣
find Δ

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Solution

ϕ1(x)=x+a1,ϕ2(x)=x2+b1x+b2,x1=2,x2=3,x3=5Δ=∣ ∣ ∣111ϕ1(x1)ϕ1(x2)ϕ1(x3)ϕ2(x1)ϕ2(x2)ϕ2(x3)∣ ∣ ∣applyingC3C3C2andC2C2C1weobtainΔ=∣ ∣ ∣100ϕ1(x1)x2x1x3x2ϕ2(x1)x22x21+b1(x2x1)x23x22+b(x3x2)∣ ∣ ∣=(x2x1)(x3x2)11x2+x1+b1x3+x2+b1=(x2x1)(x3x2)(x3x1)Δ=6

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