If $x, y$ and $z$ are all different from zero and $\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix} = 0$, then the value of $x^{-1} + y^{-1} + z^{-1}$ is

$-1$

The correct answer is Option (4) → $-1$ ##

We have, $\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix} = 0$, where $x \neq 0, y \neq 0, z \neq 0$.

Now, on expanding along $R_1$, we get

$(1+x)[(1+y)(1+z) - 1] - 1[(1+z) - 1] + 1[1 - (1+y)] = 0$

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

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

$\Rightarrow z + y + yz + xz + xy + xyz - z - y = 0$

$\Rightarrow yz + xz + xy + xyz = 0 \quad \dots(i)$

On dividing eq. (i) by $xyz$, we get

$\frac{yz}{xyz} + \frac{xz}{xyz} + \frac{xy}{xyz} + \frac{xyz}{xyz} = 0$

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

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