Question

If $x,y,z$ are all different from zero and

$\left|\begin{array}{ccc}1+x& 1& 1\\ 1& 1+y& 1\\ 1& 1& 1+z\end{array}\right|=0$, then value of

$x^{-1}+y^{-1}+z^{-1}$ is:

Answer 1

Simplifying the given determinant by using the column operation

Let, $\Delta =\left|\begin{array}{ccc}1+x& 1& 1\\ 1& 1+y& 1\\ 1& 1& 1+z\end{array}\right|$

$\Rightarrow \Delta =\left|\begin{array}{ccc}x& 0& 1\\ 0& y& 1\\ -z& -z& 1+z\end{array}\right|$

$[\text{Applying}{C}_1\to C_1-C_3\text{and}{C}_2\to C_2-C_3]$

$\Rightarrow \Delta =x\left[y\right(1+z)+z]-0+1\left(yz\right)$

[Expanding along $R_1$]

$\Rightarrow \Delta =xy+xyz+xz+yz\dots \left(1\right)$

But it’s given that $\Delta =0$

$\Rightarrow xy+xyz+xz+yz=0\dots \left(2\right)$

$\Rightarrow \frac{1}{z}+1+\frac{1}{y}+\frac{1}{x}=0$

(Dividing by $xyz$ in (2)

$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-1$

Hence correct option is 'd'.