Question

If $a,b$ and $c$ are all different from zero and $\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=0$ , then the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is

• A. -1
• B. $abc$
• C. $\frac{1}{abc}$
• D. $-a-b-c$

Answer 1

Answer: (A)

$\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=0$
$\Rightarrow (1+a)[(1+b)(1+c)-1)-1((1+c)-1)+1(1-(1+b))=0$
$\Rightarrow (1+a)(b+c+be)-c-b=0$
$\Rightarrow b+c+bc+ab+ac+abc-b-c=0$
$\Rightarrow ab+bc+ca=-abc$
$\Rightarrow \frac{1}{c}+\frac{1}{a}+\frac{1}{b}=-1$
$\therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-1$