Question

If abc $\ne 0$ and $\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=0$, then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=$

• A. $1$
• B. $2$
• C. $-1$
• D. $-3$

Answer 1

Answer: (C)

$\Delta =\left(1+a\right)\left(b+c+bc\right)-c-b=0$
$\Rightarrow abc\left[1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]=0$
Since, abc $\ne 0$
So, $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ $\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-1$