Question

The system of equation $ax+y+z=0,x+by+z=0;x+y+cz=0$ has a non-trivial solution then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$

• A. $2$
• B. $0$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

For given equation to posses non-trivial solution we must have,

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

Now use the property of $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$

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

Now expanding along first row

$\Rightarrow (a-1)\left[\right(b-1)c-(1-c\left)\right]-(1-b)[0-(1-c\left)\right]=0$

$\Rightarrow (a-1)(b-1)c-(a-1)-(1-c)+(1-b)(1-c)=0$

$\Rightarrow (1-a)(1-b)c+(1-a)(1-c)+1(1-b)(1-c)=0$

Divide both sides $(1-a)(1-b)(1-c)$

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

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

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

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