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.