Question

If a + b + c ≠ 0 and `|("a", "b","c"),("b", "c", "a"),("c", "a", "b")|` 0, then prove that a = b = c.

Answer 1

Let Δ = `|("a", "b","c"),("b", "c", "a"),("c", "a", "b")|`

[Applying R₁ → R₁ + R₂ + R₃]

Δ = `|("a" + "b" + "c", "a" + "b" + "c", "a" + "b" + "c"),("b", "c", "a"),("c", "a", "b")|`

= `("a"+ "b" + "c")|(1, 1, 1),("b", "c", "a"),("c", "a", "b")|`

[Applying C₁ → C₁ + C₃ and C₂ → C₂ – C₃]

Δ = `("a" + "b" + "c")|(0, 0,1),("b" - "a", "c" - "a", "a"),("c" - "b", "a" - "b", "b")|`

[Expanding along R₁]

= `("a" + "b" + "c")[1("b" - "a")("a" - "b") - ("c" - "a")("c" - "b")`

= `("a" + "b" + "c")("ba" - "b"^2- "a"^2 + "ab" - "c"^2 + "cb" + "ac" - "ab")`

= `-("a" + "b" + "c")("a"^2 + "b"^2 + "c"^2 - "ab" - "bc" - "ca")`

= `(-1)/2 ("a" + "b" + "c")[2"a"^2 + 2"b"^2 + 2"c"^2 - 2"ab" - 2"bc" - 2"ca"]`

= `-1/2 ("a" + "b" + "c")[("a"^2 + "b"^2 - 2"ab") + ("b"^2 + "c"^2 - 2"bc") + ("c"^2 + "a"^2 - 2"ac")]`

= `(-1)/2 ("a" + "b" + "c")[("a" - "b")^2 + ("b" - "c")^2 + ("c" - "a")^2]`

Given, Δ = 0

⇒ `(-1)/2 ("a" + "b" + "c")[("a" - "b")^2 + ("b" - "c")^2 + ("c" - "a")^2]` = 0

⇒ `("a" - "b")^2 + ("b" - "c")^2 + ("c" - "a")^2` = 0  ...[∵ a + b + c ≠ 0, given]

⇒ a – b = b – c = c – a = 0

⇒ a = b = c