If a, b, c are real numbers such that , then show that either a + b + c = 0 or a = b = c.

Let


Given that Δ = 0.


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 2(a + b + c) 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 = 2(a + b + c)(1)[(b – c)(c – b) – (c – a)(b – a)]


⇒ Δ = 2(a + b + c)(bc – b2 – c2 + cb – cb + ca + ab – a2)


∴ Δ = 2(a + b + c)(ab + bc + ca – a2 – b2 – c2)


We have Δ = 0


⇒ 2(a + b + c)(ab + bc + ca – a2 – b2 – c2) = 0


⇒ (a + b + c)(ab + bc + ca – a2 – b2 – c2) = 0


Case – I:


a + b + c = 0


Case – II:


ab + bc + ca – a2 – b2 – c2 = 0


⇒ a2 + b2 + c2 – ab – bc – ca = 0


Multiplying 2 on both sides, we have


2(a2 + b2 + c2 – ab – bc – ca) = 0


⇒ 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0


⇒ a2 – 2ab + b2 + b2 – 2bc + c2 + c2 – 2ca + a2 = 0


⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0


We know (a – b)2 ≥ 0, (b – c)2 ≥ 0, (c – a)2 ≥ 0


If the sum of three non-negative numbers is zero, then each of the numbers is zero.


⇒ (a – b)2 = 0 = (b – c)2 = (c – a)2


⇒ a – b = 0 = b – c = c – a


⇒ a = b = c


Thus, if, then either a + b + c = 0 or a = b = c.


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