Prove the following identities –
(CBSE 2014)
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 (x + y + z) 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 = (x + y + z)(1)[0 – (–(x + y + z)(x + y + z))]
⇒ Δ = (x + y + z)(x + y + z)(x + y + z)
∴ Δ = (x + y + z)3
Thus, 
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(CBSE 2014)
(CBSE 2014)
(CBSE 2014)
