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Question

Using cofactors of elements of third column evaluate Δ=∣ ∣1xyz1yzx1zxy∣ ∣

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Solution

∣ ∣1xyz1yzx1zxy∣ ∣

Cofactors of elements of third row are
C31=(1)3+1xyzyzx
C31=x2zy2z

C32=(1)3+21yz1zx
C32=(zxyz)=yzzx

C33=(1)3+31x1y
C33=yx

Now, Δ=1C31+zC32+xyC33
Δ=x2zy2z+z(yzzx)+xy(yx)
Δ=x2zy2z+yz2xz2+xy2x2y

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