Question

If x, y, z are 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

$$\begin{gathered} \left|\begin{array}{ccc}1+x& 1& 1\\ 1& 1+y& 1\\ 1& 1& 1+z\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}x& 0& -z\\ 0& y& -z\\ 1& 1& 1+z\end{array}\right|=0\left[\mathrm{Applying}{R}_2\to R_2-R_{3}and{R}_1\to R_1-R_3\right] \\ \Rightarrow x\left[y\left(1+z\right)+z\right]+1\left(yz\right)=0\left[\mathrm{Expanding}\mathrm{along}\mathrm{first}\mathrm{column}\right] \\ \Rightarrow x\left[y+yz+z\right]+yz=0 \\ \Rightarrow xy+xyz+xz+yz=0 \\ \Rightarrow xy+yz+zx=-xyz \\ \Rightarrow \frac{xy}{xyz}+\frac{yz}{xyz}+\frac{zx}{xyz}=-\frac{xyz}{xyz} \\ \Rightarrow \frac{1}{z}+\frac{1}{x}+\frac{1}{y}=-1 \\ \Rightarrow x^{-1}+y^{-1}+z^{-1}=-1 \end{gathered}$$

Hence, the correct option is (d).