Question

The determinant $\left|\begin{array}{ccc}1& ab& \frac{1}{a}+\frac{1}{b}\\ 1& bc& \frac{1}{b}+\frac{1}{c}\\ 1& ca& \frac{1}{c}+\frac{1}{a}\end{array}\right|$ is equal to

• A. 0
• B. bc + ca + ab
• C. $a^3+b^3+c^3$
• D. None of these

Answer 1

The correct option is A

0

$\Delta =\left|\begin{array}{ccc}1& ab& \frac{1}{a}+\frac{1}{b}\\ 1& bc& \frac{1}{b}+\frac{1}{c}\\ 1& ca& \frac{1}{c}+\frac{1}{a}\end{array}\right|\Delta =abc\left|\begin{array}{ccc}1& \frac{1}{c}& \frac{1}{a}+\frac{1}{b}\\ 1& \frac{1}{a}& \frac{1}{b}+\frac{1}{c}\\ 1& \frac{1}{b}& \frac{1}{c}+\frac{1}{a}\end{array}\right|\Delta =abc\left|\begin{array}{ccc}1& \frac{1}{c}& (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\\ 1& \frac{1}{a}& (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\\ 1& \frac{1}{b}& (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\end{array}\right|(C_3\to C_3+C_2)\Delta =abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\left|\begin{array}{ccc}1& \frac{1}{c}& 1\\ 1& \frac{1}{a}& 1\\ 1& \frac{1}{b}& 1\end{array}\right|=0$