Question

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

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

Answer 1

The correct option is D $-1$

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

$R_1\to R_1-R_2{R}_3\to R_3-R_2$

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

$=a(c+bc+b)+bc=0$

$=ab+bc+ac+abc=0$

$=ab+bc+ac=-abc$

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

Option D