If the system of linear equations $$a(y+z)-x=0$$ $$b(z+x)-y=0$$$$c(x+y)-z=0$$ has a non-trivial solution $(a,b,c \neq -1)$,then show that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2$$
MY ATTEMPT:
I took this determinant and put it to $0$.
$\begin{vmatrix} -1 & a & a \\ b & -1 & b \\ c & c & -1 \end{vmatrix}=0$
I'm getting $2abc+ab+bc+ca-1=0$ but that is not what is asked for! Someone help me out !
You are already finished, because $$ \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}-2=\frac{ - 2abc - ab - ac - bc + 1}{abc + ab + ac + a + bc + b + c + 1}=0. $$