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Question

Usng properties of determinants , prove that ∣ ∣111+3x1+3y1111+3z1∣ ∣=9(3xyz+xy+yz+zx)

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Solution

LHS : Let Δ=∣ ∣111+3x1+3y1111+3z1∣ ∣By C1C1C2,C2C2C3Δ=∣ ∣03x1+3x3y013z3z1∣ ∣ Expanding along R1

Δ=9{0+x(y+z)+(1+3x)(yz0)}

Δ=9{3xyz+xy+yz+zx} =RHS


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