Question

If $a,b,c$ are non-zero real numbers, then
$\Delta =\left|\begin{array}{ccc}1& ab& \frac{1}{a}+\frac{1}{b}\\ 1& bc& \frac{1}{b}+\frac{1}{c}\\ 1& ca& \frac{1}{c}+\frac{1}{a}\end{array}\right|$ is

• A. $0$
• B. $bc+ca+ab$
• C. $a^{-1}+b^{-1}+c^{-1}$
• D. none of these

Answer 1

The correct option is A $0$

$\Delta =\left|\begin{array}{ccc}1& ab& \frac{1}{a}+\frac{1}{b}\\ 1& bc& \frac{1}{b}+\frac{1}{c}\\ 1& ca& \frac{1}{c}+\frac{1}{a}\end{array}\right|$

Taking $abc$ common from $R_2$, we get

$\Delta =abc\left|\begin{array}{ccc}1& \frac{1}{c}& \frac{1}{a}+\frac{1}{b}\\ 1& \frac{1}{a}& \frac{1}{b}+\frac{1}{c}\\ 1& \frac{1}{b}& \frac{1}{c}+\frac{1}{a}\end{array}\right|$

Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$

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

Hence, option 'A' is correct.