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

Question

If a, b, c are real numbers such that \(\begin{vmatrix} b+c& c+a & a+b \\[0.3em] c+a& a+b & b+c \\[0.3em] a+b &b+c &c+a \end{vmatrix}\) = 0, then show that either a + b + c = 0 or a = b = c.

Answer 1

Let Δ = \(\begin{vmatrix} b+c& c+a & a+b \\[0.3em] c+a& a+b & b+c \\[0.3em] a+b &b+c &c+a \end{vmatrix}\)

Given that,

Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, 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  \(\begin{vmatrix} b+c& c+a & a+b \\[0.3em] c+a& a+b & b+c \\[0.3em] a+b &b+c &c+a \end{vmatrix}\) = 0, then either a + b + c = 0 

or a = b = c.