Question

If the system of equations $ax+by+c=0;bx+cy+a=0;cx+ay+b=0$ has a non-trivial solutions, the system of equation,

$(b+c)x+(c+a)y+(a+b)z=0;(c+a)x+(a+b)y+(b+c)z=0;(a+b)x+(b+c)y+(c+a)z=0$

has

• A. Only one solution
• B. No solution
• C. Infinite number of solution
• D. None of these

Answer 1

The correct option is C Infinite number of solution

For existence of a non-trivial solution of the

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

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

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

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

$\Rightarrow \left|\begin{array}{cc}c-b& a-c\\ a-b& b-c\end{array}\right|=0$ or

$(a+b+c)=0$ ...(1)

The second system will have a non-trivial solution if we can prove that

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

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

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

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

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

$\therefore$ The second system will have a non-trivial solution.