Question

If a,b, and c are real numbers and determinant \(\Delta = \begin{vmatrix} b+c &c+a  &a+b \\   c+a&a+b  &b+c \\   a+b&b+c  &c+a  \end{vmatrix}\)
Show that either a+b+c=0 or a=b=c. 

Answer 1

\(\Delta = \begin{vmatrix} b+c &c+a  &a+b \\   c+a&a+b  &b+c \\   a+b&b+c  &c+a  \end{vmatrix}\)
Applying R₁\(\rightarrow\)R₁+R₂+R₃, we have,
\(\Delta = \begin{vmatrix} 2(a+b+c) &2(a+b+c)   &2(a+b+c)  \\   c+a&a+b  &b+c \\   a+b&b+c  &c+a  \end{vmatrix}\)
=2(a+b+c)\(\begin{vmatrix}  1&1  &1 \\   c+a&a+b &b+c \\   a+b&b+c  &c+a  \end{vmatrix}\)
Applying C₂\(\rightarrow\)C₂-C₁ and C₃\(\rightarrow\)C₃-C₁,we have,
Δ=2(a+b+c)\(\begin{vmatrix}  1&1  &1 \\   c+a&a+b &b+c \\   a+b&b+c  &c+a  \end{vmatrix}\)

Expanding along R₁,we have:
Δ=2(a+b+c)(1)[(b-c)(c-b)-(b-a)(c-a)]
=2(a+b+c)[-b²-c²+2bc-bc+ba+ac-a²]
=2(a+b+c)[ab+bc+ca-a²-b²-c²]

It is given that Δ=0.
(a+b+c)[ab+bc+ca-a²-b²-c²]=0
⇒ Either a+b+c=0,or ab+bc+ca-a²-b²-c²=0.

Now,
ab+bc+ca-a²-b²-c²=0.
⇒ -2ab-2bc-2ca+2a²+2b²+2c²=0
⇒ (a-b)²+(b-c)²+(c-a)²=0
⇒ (a-b)²=(b-c)²=(c-a)²=0     [(a-b)²,(b-c)²,(c-a)² are non-negative]
⇒ (a-b)=(b-c)=(c-a)=0
⇒ a=b=c

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