Question

If the equations $(b+c)x+(c+a)y+(a+b)z=0;$ $cx+ay+bz=0$; $ax+by+cz=0$ have non-zero solutions then a relation among a, b, c is

• A. $a=b=c=0$
• B. $a=2b=3c$
• C. $a=b=c$
• D. $a+b+c=0$

Answer 1

Answer: (C)

$a=b=c$

For non- zero solution
$\left|\begin{array}{ccc}(b+c)& (c+a)& (a+b)\\ c& a& b\\ a& b& c\end{array}\right|$ should be equal to $0$
$R_1\to R_1-R_2$
$\left|\begin{array}{ccc}b& c& a\\ c& a& b\\ a& b& c\end{array}\right|=0$
$R_1\to R_1+R_2+R_3$
$(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ c& a& b\\ a& b& c\end{array}\right|=0$
$(a+b+c)(ac-b^2+ab-c^2+bc-a^2)=0$
$\frac{1}{2}(a+b+c)(2a^2+2b^2+2c^2-2ac-2bc-2ab)=0$
$\frac{1}{2}(a+b+c)(a^2+b^2-2ab+a^2+c^2-2ac+b^2+c^2-2bc)=0$

$\frac{1}{2}(a+b+c)\left(\right(b-a{)}^2+(a-c{)}^2+(c-b{)}^2)=0$

$\therefore a=b=c$