question_answer

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)        \[3\]

Correct Answer: B

Solution :

for non-zero solution

\[\Rightarrow \,abc-a-b-c-2=0\]

\[\Rightarrow \,abc=a+b+c+2\]

Now, \[\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\]\[=\frac{3+2(a+b+c)+(ab+bc+ac)}{1+(a+b+c)+(ab+bc+ac)+abc}\]

\[=\frac{3+2(a+b+c)+(ab+bc+ac)}{1+2(a+b+c)+2+ab+bc+ac}=1\]