Question

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$, then $\left|\begin{array}{c}1+a\\ 1\\ 1\end{array}\begin{array}{c}1\\ 1+b\\ 1\end{array}\begin{array}{c}1\\ 1\\ 1+c\end{array}\right|$ is equal to

• A. $0$
• B. $abc$
• C. $-abc$
• D. None of these

Answer 1

Answer: (B)

$abc$

Let $D=$ $\left|\begin{array}{c}1+a\\ 1\\ 1\end{array}\begin{array}{c}1\\ 1+b\\ 1\end{array}\begin{array}{c}1\\ 1\\ 1+c\end{array}\right|$
$\therefore D=$ $(1+a)\left[\right(1+b\left)\right(1+c)-1]-1\left[\right(1+c)-1]+1[1-1-b]$
$\therefore D=ac+ab+bc+abc$
Since, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
$\therefore ab+bc+ac=0$
Hence, $D=abc$