Question

If the system of equations
$ax+ay-z=0$
$bx-y+bz=0$ and
$-x+cy+cz=0$
(where $a,b,c\ne -1$ ) has a non trivial solution, then the value of $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$ is

Answer 1

Given that,
$ax+ay-z=0$
$bx-y+bz=0$
$-x+cy+cz=0$ has non trivial solution
$\Rightarrow \Delta =0$
$\Rightarrow \left|\begin{array}{ccc}a& a& -1\\ b& -1& b\\ -1& c& c\end{array}\right|=0$

Applying $C_1\to C_1-C_3;C_2\to C_2-C_3$
$\Rightarrow \left|\begin{array}{ccc}a+1& a+1& -1\\ 0& -(1+b)& b\\ -(1+c)& 0& c\end{array}\right|=0$
$\Rightarrow -(a+1)(b+1)c-(1+c)\left[\right(a+1)b-(b+1\left)\right]=0$
$\Rightarrow (c+1)(b+1)=(a+1)(b+1)c+(a+1)(c+1)b$
On dividing both sides by $(a+1)(b+1)(c+1),$ we have
$\frac{1}{a+1}=\frac{c}{c+1}+\frac{b}{b+1}$
$\Rightarrow \frac{1}{a+1}=1-\frac{1}{c+1}+1-\frac{1}{b+1}$
$\Rightarrow \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2$