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Question
Using Cofactors of elements of third column, evaluate Δ = `|(1,x,yz),(1,y,zx),(1,z,xy)|`.
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Solution
Δ = `|(1,x,yz),(1,y,zx),(1,z,xy)|`
A13 = `-1^(1 + 3) |(1,y),(1,z)|`
= (1) (y − z)
= (y − z)
A23 = `-1^ (2 + 3) |(1,x),(1,z)|`
= (−1) (x − z)
= (z − x)
A33 = `-1^(1 + 3) |(1,x),(1,y)|`
= (1) (x − y)
= (x − y)
Δ = a13A13 + a23A23 + a33A33
= yz(y − z) + zx(z − x) + xy(x − y)
= y2z − yz2 + z2x − zx2 + x2y − xy2
= x2y + y2z + z2x − xy2 − yz2 − zx2
= (x2y − xy2) + (y2z − yz2) + (z2x − zx2)
= xy(x − y) + yz(y − z) + zx(z − x)
= (x − y)xy + (y − z)yz + (z − x)zx
= (x − y)(y − z)(z − x)
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