Q40 of 256 Page 7

Prove the following identities –

= abc(a2 + b2 + c2)3

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 Ri→ Ri + kRj or Ci→ Ci + kCj.


Applying R1→ R1 – R2, we get




Taking the term (a2 + b2 + c2) common from R1, we get



Applying R2→ R2 – R3, we get




Taking the term (a2 + b2 + c2) common from R2, we get



Applying C2→ C2 + C1, we get




Expanding the determinant along R1, we have


Δ = (abc)(a2 + b2 + c2)2(–1)[(a2 + b2 – c2) – (2b2 + 2a2)]


⇒ Δ = (abc)(a2 + b2 + c2)2[–(a2 + b2 – c2) + (2b2 + 2a2)]


⇒ Δ = (abc)(a2 + b2 + c2)2[–a2 – b2 + c2 + 2b2 + 2a2]


⇒ Δ = (abc)(a2 + b2 + c2)2[a2 + b2 + c2]


∴ Δ = (abc)(a2 + b2 + c2)3


Thus,


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