Prove the following identities –

Let

Taking a, b and c common from C₁, C₂ and C₃, we get

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 C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Multiplying a, b and c to R₁, R₂ and R₃, we get

Applying R₁→ R₁ + R₂, we get

Applying R₁→ R₁ + R₃, we get

Expanding the determinant along R₁, we have

Δ = (1 + a² + b² + c²)[(1)(1) – (0)(0)] – 0 + 0

⇒ Δ = (1 + a² + b² + c²)(1)

∴ Δ = 1 + a² + b² + c²

Thus,