Prove the following identities –

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ – R₂, we get

Applying R₁→ R₁ – R₃, we get

Taking the term (–2x) common from R₁, we get

Applying C₂→ C₂ – C₃, we get

Expanding the determinant along R₁, we have

Δ = (–2x)[(z)(–y) – (y)(z)]

⇒ Δ = (–2x)(–yz –yz)

⇒ Δ = (–2x)(–2yz)

∴ Δ = 4xyz

Thus,