Question

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

• A. $0$
• B. $abc$
• C. $-abc$
• D. $2abc$

Answer 1

The correct option is B $abc$

$\begin{array}{c}{R}_1\to R_1-R_2\\ R_2\to R_2-R_3\end{array}$ $\left|\begin{array}{ccc}a& -b& 0\\ 0& b& -c\\ 1& 1& 1+c\end{array}\right|$

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

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

$=abc+ab+bc+ac$

$=abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=abc$