Prove the following identities –
Let
Taking a2, b2 and c2 common from C1, C2 and C3, we get
Taking a, b and c common from R1, R2 and R3, we get
Recall that the value of a determinant remains same if we apply the operation Ri→ Ri + kRj or Ci→ Ci + kCj.
Applying C2→ C2 – C3, we get
Expanding the determinant along R1, we have
Δ = (a3b3c3)[0 – 0 + 1(1)(1) – (1)(–1)]
⇒ Δ = (a3b3c3)[1 + 1]
∴ Δ = 2a3b3c3
Thus,