Question

If a,b and c are real numbers,and determinant Δ=|b+c c+a a+b c+a a+b b+c a+b b+c c+a|=0.
Show that either a+b+c=0 or a=b=c. 

Answer 1

Δ=|b+c c+a a+b c+a a+b b+c a+b b+c c+a|
Applying R1→R1+R2+R3,we have,
Δ=|2(a+b+c) 2(a+b+c) 2(a+b+c) c+a a+b b+c a+b b+c c+a|
=2(a+b+c)|111 c+a a+b b+c a+b b+c c+a|
Applying C2→C2-C1 and C3→C3-C1,we have,
Δ=2(a+b+c)|100 c+a a+b b+c a+b b+c c+a|

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.