Question

If a, b and c are real numbers and $\Delta =\left|\begin{array}{ccc}b+c& c+a& a+b\\ c+a& a+b& b+c\\ a+b& b+c& c+a\end{array}\right|=0$, Show that either a+b+c=0 or a=b=c.

Answer 1

Given, $\Delta =\left|\begin{array}{ccc}b+c& c+a& a+b\\ c+a& a+b& b+c\\ a+b& b+c& c+a\end{array}\right|=\left|\begin{array}{ccc}2(a+b+c)& c+a& a+b\\ 2(a+b+c)& a+b& b+c\\ 2(a+b+c)& b+c& c+a\end{array}\right|$
(using $C_1\to C_1+C_2+C_3$)
$=2(a+b+c)\left|\begin{array}{ccc}1& c+a& a+b\\ 1& a+b& b+c\\ 1& b+c& a+b\end{array}\right|$ (Taking out factor 2(a+b+c) from $C_1$)
$=2(a+b+c)\left|\begin{array}{ccc}1& c+a& a+b\\ 0& b-c& c-a\\ 0& b-a& c-b\end{array}\right|$ (using $R_2\to R_2-R_1,R_3\to R_3-R_1$)
By expanding along $C_1$, we get
$=2\left(a\right)+b+c\left[\right(b-c\left)\right(c-b)-(b-a\left)\right(c-a\left)\right]=2(a+b+c)[bc-c^2-b^2+bc-bc+ac+ab-a^2]=2(a+b+c)[ab+bc+ca-a^2-b^2-c^2]$
It is given that $\Delta =0$
$\therefore (a+b+c)\times 2[ab+bc+ca-a^2-b^2-c^2]=0$
$\Rightarrow$ Either $(a+b+c)=0or2(ab+bc+ca-a^2-b^2-c^2)=0$
If $2[ab+bc+ca-a^2-b^2-c^2]=0$
$\Rightarrow -(2a^2+2b^2+2c^2-2ab-2bc-2ca)=0$
[Taking negative sign common]
$\Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0\Rightarrow (a^2+b^2-2ab)+(b^2+c^2-2bc)+(c^2+a^2-2ca)=0\Rightarrow (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0[\because (x-y{)}^2=x^2+y^2-2xy]$
$\Rightarrow (a-b{)}^2=(b-c{)}^2(c-a{)}^2=0$ [$\because (a-b{)}^2,(b-c{)}^2,(c-a{)}^2$ are positive]
$\Rightarrow a=b=c$
Hence, if $\Delta =0$, then either $a+b+c=0ora=b=c$.