Question

If a, b, c all are non-zero and unequal and $\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=0$, then $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is equal to

• A. $abc$
• B. $\frac{1}{abc}$
• C. $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
• D. $0$

Answer 1

The correct option is D $0$

$a\ne b\ne c\ne 0$

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

$abc\left|\begin{array}{ccc}\frac{1}{a}+1& \frac{1}{a}& \frac{1}{a}\\ \frac{1}{b}& \frac{1}{b}+1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|=0$
$R_1\to R_1+R_2+R_3$

$\left(abc\right)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\left|\begin{array}{ccc}1& 1& 1\\ \frac{1}{b}& \frac{1}{b}-1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|=0$
$C_1\to C_1-C_2;C_2\to C_2-C_3$

$\left(abc\right)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\left|\begin{array}{ccc}0& 0& 1\\ -1& 1& \frac{1}{b}\\ 0& -1& \frac{1}{c}+1\end{array}\right|=0$
$\left(abc\right)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=0$
$\because abc\ne 0$ $\Rightarrow$ $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
Ans: D