Question

If a system of equation -ax+y+z=0, x-by+z=0, x+y-cz=0, $(a,b,c\ne -1)$ has a non-zero solution then $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

Given equation are,

$-ax+y+z=0,x-by+z=0,x=y-cz=0$

$⟹a=\frac{y+z}{x},b=\frac{x+z}{y},c=\frac{x+y}{z}$

Let calculate

$\frac{a}{1+a}=\frac{\frac{y+z}{x}}{1+\frac{y+z}{x}}$

$=\frac{y+z}{x+y+z}-\left(1\right)$

Similarly

$\frac{b}{1+b}=\frac{x+z}{x+y+z}-\left(2\right)$

Also

$\frac{c}{1+c}=\frac{x+y}{x+y+z}-\left(3\right)$

Adding qs.(1),(2) and (3)-

$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=\frac{2(x+y+z)}{x+y+z}$

Subtracting $3$ on Both sides,

$⟹(1-\frac{a}{a+1})+(1-\frac{b}{b+1})+(1-\frac{c}{1+c})=3-2$

$⟹\frac{1}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1$