Question

If the system of equations $ax+y+z=0;x+by+z=0$ and $x+y+cz=0$ has a non-trivial solution, then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is

• A. $1$
• B. $2$
• C. $-1$
• D. $0$

Answer 1

The correct option is A $1$

Given equations are
$ax+y+z=0;x+by+z=0$ and $x+y+cz=0$
They have non trivial solutions, so
$\left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0$
Using row operations
$R_1\to R_1-R_2$ and $R_2\to R_2-R_3,$ we get
$\left|\begin{array}{ccc}a-1& 1-b& 0\\ 0& b-1& 1-c\\ 1& 1& c\end{array}\right|=0\Rightarrow (a-1)\left[\right(b-1)c-(1-c\left)\right]-(1-b)[0-(1-c\left)\right]=0\Rightarrow (a-1)(b-1)c-(a-1)(1-c)+(1-b)(1-c)=0\Rightarrow (1-a)(1-b)c+(1-a)(1-c)+(1-b)(1-c)=0$

Dividing both sides by $(1-a)(1-b)(1-c),$ we get

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