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 C2→ C2 – pC1, we get
Applying C3→ C3 – qC1, we get
Applying C3→ C3 – pC2, we get
Applying C2→ C2 – C1, we get
Applying C3→ C3 – C1, we get
Expanding the determinant along R1, we have
Δ = 1[(1)(7) – (3)(2)] – 0 + 0
∴ Δ = 7 – 6 = 1
Thus,