Mathematics › Determinants › System of Linear Equations
If α+β+γ=2π, then the system of equations
x+cosγy+cosβz=0
cosγx+y+cosαz=0
cosβx+cosαy+z=0
has :
Given, α+β+γ=2π
Now, Δ=1cosγcosβcosγ1cosαcosβcosα1
By expanding along the first column, we get
∆=11-cos2α-cosγcosγ-cosα.cosβ+cosβcosα.cosγ-cosβ
∆=1-cos2α-cos2γ+cosγ.cosα.cosβ+cosβ.cosα.cosγ-cos2β
=1-cos2α-cos2β-cos2γ+2cosα.cosβ.cosγ
=sin2α-cos2β-cosγ(cosγ-2cosα.cosβ)
=-cos(α+β).cos(α-β)-cosγcos2π-(α+β)-2cosα.cosβ
=-cos(2π-γ)cos(α-β)-cosγcos(α+β)-2cosα.cosβ
=-cos(2π-γ).cos(α-β)-cosγcosα.cosβ-sinα.sinβ-2cosα.cosβ
=-cosγ.cos(α-β)+cosγ.cos(α-β)
=0
So, ∆=0
Hence, the system of equations has infinitely many solutions.
Asked in: JEE Main 2021 (31 Aug Shift 2)
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