Answer :

Given, Δ =

Applying Elementary transformations, we get

R_{1}→ R_{1} + R_{2} + R_{3} we have,

Δ = 2 (a + b + c)

Now applying C_{2}→ C_{2} – C_{1} and C_{3}→ C_{3} – C_{1}

Δ = 2 (a + b + c)

Expanding along R_{1}

Δ = 2 (a + b + c) [1 {(b – c)(c – b) – (b – a)(c – a)} – 0 + 0]

Δ = 2 (a + b + c) [- b^{2} – c^{2} + 2bc – bc + ba + ac – a^{2}]

Δ = 2 (a + b + c) [ab + bc + ca – a^{2} – b^{2} – c^{2}]

Given that Δ = 0

∴ 2 (a + b + c) [ab + bc + ca – a^{2} – b^{2} – c^{2}] = 0

⇒ Either a + b + c = 0, or ab + bc + ca – a^{2} – b^{2} – c^{2} = 0

Now

ab + bc + ca – a^{2} – b^{2} – c^{2} = 0

Multiplying both sides by -2

⇒ - 2ab – 2bc – 2ca + 2a^{2} + 2b^{2} + 2c^{2} = 0

⇒ a^{2} – 2ab + b^{2} + b^{2} – 2bc + c^{2} + c^{2} – 2ca + a^{2} = 0

⇒ (a – b)^{2} + (b – c)^{2} + (c – a)^{2} = 0

Since, (a – b)^{2}, (b – c)^{2}, (c – a)^{2} are non-negative

∴ (a – b)^{2} = (b – c)^{2} = (c – a)^{2}

⇒ (a – b) = (b – c) = (c – a)

⇒ a = b = c

Hence, if ∆ = 0, then either a + b + c = 0 or a = b = c

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Using properties Mathematics - Board Papers

Using properties Mathematics - Board Papers

Using properties Mathematics - Board Papers

Prove the followiMathematics - Board Papers