Question

If the system of equations $ax+by+c=0$ $bx+cy+a=0$ ,$cx+ay+b=0$ has a solution then the system of equations $(b+c)x+(c+a)y+(a+b)z=0$ ,$(c+a)x+(a+b)y+(b+c)z=0$ , $(a+b)x+(b+c)y+(c+a)z=0$ has

• A. only one solution
• B. no solution
• C. infinite number of solutions
• D. none of these

Answer 1

Answer: (C)

infinite number of solutions

$ax+by+c=0,bx+cy+a=0,cx+ay+b=0$
Gives
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$
Now $(b+c)x+(c+a)y+(a+b)z=0,(c+a)x+(a+b)y+(b+c)z=0,(a+b)x+(b+c)y+(c+a)z=0$
gives
$\left|\begin{array}{ccc}b+c& c+a& a+b\\ c+a& a+b& b+c\\ a+b& b+c& c+a\end{array}\right|=\left|\begin{array}{ccc}b& c& a\\ c& a& b\\ a& b& c\end{array}\right|+\left|\begin{array}{ccc}c& a& b\\ a& b& c\\ b& c& a\end{array}\right|=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|+\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=2\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$