Question

If x, y and 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 the value of $x^{-1}+y^{-1}+z^{-1}$ is

(a) $xyz$
(b) $x^{-1}{y}^{-1}{z}^{-1}$
(c) $-x-y-z$
(d) $-1$

Answer 1

(d) We have, $\left|\begin{array}{ccc}1+x& 1& 1\\ 1& 1+y& 1\\ 1& 1& 1+z\end{array}\right|=0$
Applying $C_1\to C_1-C_3$ and $C_2\to C_2-C_3$
$\Rightarrow \left|\begin{array}{ccc}x& 0& 1\\ 0& y& 1\\ -z& -z& 1+z\end{array}\right|=0$
Expanding along $R_1$,
$x\left[y\right(1+z)+z]-0+1\left(yz\right)=0\Rightarrow x(y+yz+z)+yz=0\Rightarrow xy+xyz+xz+yz=0$
$\Rightarrow \frac{xy}{xyz}+\frac{xyz}{xyz}+\frac{xz}{xyz}+\frac{yz}{xyz}=0$ [on dividing (xyz) from both sides]
$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1=0\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-1\therefore x^{-1}+y^{-1}+z^{-1}=-1$