To evaluate a determinant using cofactors, Let

B =

Expanding along Row 1

B =

B = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃

[Where A_ij represents cofactors of a_ij of determinant B.]

B = Sum of product of elements of R₁ with their corresponding cofactors

Similarly, the determinant can be solved by expanding along column

So, B = sum of product of elements of any row or column with their corresponding cofactors

Cofactors of third column

A₁₃ = (-1)¹⁺³ × M₁₃ = 1 × = 1 × (1 × z – 1 × y) = (z – y)

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × = (-1) × (1 × z – 1 × x) = - (z - x) = (x - z)

A₃₃ = (-1)³⁺³ × M₃₃ = 1 × = 1 × (1 × y – 1 × x) = (y – x)

[Where A_ij = (-1)^i+j × M_ij, M_ij = Minor of i^th row & j^th column]

Therefore,

Δ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃

Δ = yz (z - y) + zx (x - z) + xy (y - x) = z [y (z - y) + x (x - z)] + xy (y - x)

Δ = z (yz - y² + x² - xz) + xy (y - x) = z [(yz - xz) + (x² - y²)] + xy (y - x)

Δ = z [z × (y - x) + (x + y) × (x - y)] + xy (y - x)

Δ = z × (y - x) × (z – x - y) + xy (y - x)

Δ = (y - x) × (z² – xz – yz + xy)

Δ = (y - x) × [z (z - x) – y (z - x)] = (y - x) × (z - y) × (z - x)

Δ = (x - y) (y - z) (z - x)

Ans: Δ = (x - y) (y - z) (z - x)