Question

$a,b,c$ are distinct real numbers, not equal to one. If $ax+y+z=0,x+by+z=0$ and $x+y+cz=0$ have a non-trivial solution, then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to

• A. $-1$
• B. $1$
• C. zero
• D. none of these

Answer 1

Answer: (B)

$1$

sincethe system has non trivial solution

$\therefore \left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0$

applying $R_1\to R_1-R_2,R_2\to R_2-R_3$ we get

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

dividing throughout by $(1-a)(1-b)(1-c)$ we get

$\frac{c}{1-c}+\frac{1}{1-c}+\frac{1}{1-b}=0\Rightarrow -1+\frac{1}{1-c}+\frac{1}{1-b}+\frac{1}{1-a}=0\Rightarrow \frac{1}{1-c}+\frac{1}{1-a}+\frac{1}{1-b}=1$