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

$\Delta =\left|\begin{array}{ccc}b+c& c+a& a+b\\ c+a& a+b& b+c\\ a+b& b+c& c+a\end{array}\right|=0$

$C_1\to C_1+C_2+C_3$

$\left|\begin{array}{ccc}2(a+b+c)& c+a& a+b\\ 2(a+b+c)& a+b& b+c\\ 2(a+b+c)& b+c& c+a\end{array}\right|=0$

Taking $2(a+b+c)$ common from $C_1$

$2(a+b+c)\left|\begin{array}{ccc}1& c+a& a+b\\ 1& a+b& b+c\\ 1& b+c& c+a\end{array}\right|=0$

$R_1\to R_1-R_2,R_2\to R_2-R_3$

$2(a+b+c)\left|\begin{array}{ccc}0& c-b& a-c\\ 0& a-c& b-a\\ 1& b+c& c+a\end{array}\right|=0$

On expanding along first column, we get

$2(a+b+c)\left[\right(c-b\left)\right(b-a)-(a-c{)}^2]=0$

$\Rightarrow 2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$

$\Rightarrow (a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)=0$

$\Rightarrow (a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$

$\Rightarrow (a+b+c)=0$ or $(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$

$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$ , which is possible only when $a=b=c$

Hence, $\Delta =0$

$\Rightarrow a+b+c=0$ or $a=b=c$