Question

If $a,b$ and $c$ are real numbers, and

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

show that either

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

Answer 1

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

Taking $L.H.S$

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

Applying $R_1\to R_1+R_2+R_3$

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

Taking common $2(a+b+c)$ from $R_1$

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

Applying $C_2\to C_2-C_1$

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

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

Applying $C_3\to C_3-C_1$

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

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

Expanding determinant along $R_1$

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

$△=2(a+b+c)\left(\right(c-b\left)\right(b-c)-(b-a\left)\right(c-a\left)\right)$

$△=2(a+b+c)\left({}_{(bc-ab-ac+a^2)}^{-(c-b)(c-b)}\right)$

$△=2(a+b+c)\left({}_{-bc+ab+ac-a^2}^{-c^2-b^2-2cb}\right)$


$△=2(a+b+c)\left({}_{-bc+ab+ac-a^2}^{--c^2-b^2+2cb}\right)$


$△=2(a+b+c)\left({}_{+ab+bc+ac}^{-(a^2-b^2-c^2)}\right)$

$\therefore △=2(a+b+c)\left({}_{+ab+bc+ac}^{-(a^2+b^2+c^2)}\right)$

Given:$△=0$

$\Rightarrow 2(a+b+c)\left({}_{+ab+bc+ac}^{-(a^2+b^2+c^2)}\right)=0$

$\Rightarrow (a+b+c)\left({}_{+ab+bc+ac}^{-(a^2+b^2+c^2)}\right)=0$

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

$[-(a^2)+b^2+c^2+ab+bc+ac]=0$

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

Multiplying & Dividing by $2$

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

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

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

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

$\Rightarrow a-b=0,b-c=0$ or $c-a=0$

$a=b,b=c$ or $c=a$

Thus, $a=b=c$

So, either $(a+b+c)=0$or $a=b=c$

Hence Proved.