Question

If x, y, z are 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

Options

• xyz

•  x⁻¹ y⁻¹ z⁻¹

• − x − y − z

• − 1

Answer 1

\[\begin{vmatrix}1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z\end{vmatrix} = 0\]
\[ \Rightarrow \begin{vmatrix}x & 0 & - z \\ 0 & y & - z \\ 1 & 1 & 1 + z\end{vmatrix} = 0 \left[\text{ Applying }R_2 \to R_2 - R_3\text{ and }R_1 \to R_1 - R_3 \right]\]
\[ \Rightarrow x\left[ y\left( 1 + z \right) + z \right] + 1\left( yz \right) = 0 \left[\text{ Expanding along first column }\right]\]
\[ \Rightarrow x\left[ y + yz + z \right] + yz = 0\]
\[ \Rightarrow xy + xyz + xz + yz = 0\]
\[ \Rightarrow xy + yz + zx = - xyz\]
\[ \Rightarrow \frac{xy}{xyz} + \frac{yz}{xyz} + \frac{zx}{xyz} = - \frac{xyz}{xyz}\]
\[ \Rightarrow \frac{1}{z} + \frac{1}{x} + \frac{1}{y} = - 1\]
\[ \Rightarrow x^{- 1} + y^{- 1} + z^{- 1} = - 1\]