Prove the following identities –
Let
Recall that the value of a determinant remains same if we apply the operation Ri→ Ri + kRj or Ci→ Ci + kCj.
Applying R1→ R1 + R2, we get
Applying R1→ R1 + R3, we get
Taking the term (3 + a) common from R1, we get
Applying C2→ C2 – C1, we get
Applying C3→ C3 – C1, we get
Expanding the determinant along R1, we have
Δ = (3 + a)(1)[(a)(a) – 0]
⇒ Δ = (3 + a)(a2)
∴ Δ = a3 + 3a2
Thus,