Question

If $a, b$ and $c$ are all different from zero and $\begin{vmatrix}1+a&1&1\\ 1&1+b&1\\ 1&1&1+c\end{vmatrix} = 0$ , 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)

$\begin{vmatrix}1+a&1&1\\ 1&1+b&1\\ 1&1&1+c\end{vmatrix} = 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$