Q67 / 180 · KCET 2016 · Mathematics
If \( x y z \) are all different and not equal to zero and \( \left|\begin{array}{cccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0 \)
then the value of \( x^{-1}+y^{-1}+z^{-1} \) is equal to
then the value of \( x^{-1}+y^{-1}+z^{-1} \) is equal to
- A\( x y z \)
- B\( x^{-1} y^{-1} z^{-1} \)
- C\( -x-y-z \)
- Correct answerD\( -1 \)
Solution
Given that,
\[ \begin{array}{l} \left|\begin{array}{ccc} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{array}\right|=0 \\ R_{1} \rightarrow R_{1}-R_{2} \text { and } R_{2} \rightarrow R_{2}-R_{3} \\ \left|\begin{array}{ccc} x & -y & 0 \\ 0 & y & -z \\ 1 & 1 & 1+z \end{array}\right|=0 \\ \Rightarrow x[y(1+z)+z]+y(z)=0 \\ \Rightarrow x y+y z+z x+x y z=0 \end{array} \]
Divide by xyz both side, we get
\[ \begin{array}{l} \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1=0 \\ \Rightarrow x^{-1}+y^{-1}+z^{-1}=-1 \end{array} \]
\[ \begin{array}{l} \left|\begin{array}{ccc} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{array}\right|=0 \\ R_{1} \rightarrow R_{1}-R_{2} \text { and } R_{2} \rightarrow R_{2}-R_{3} \\ \left|\begin{array}{ccc} x & -y & 0 \\ 0 & y & -z \\ 1 & 1 & 1+z \end{array}\right|=0 \\ \Rightarrow x[y(1+z)+z]+y(z)=0 \\ \Rightarrow x y+y z+z x+x y z=0 \end{array} \]
Divide by xyz both side, we get
\[ \begin{array}{l} \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1=0 \\ \Rightarrow x^{-1}+y^{-1}+z^{-1}=-1 \end{array} \]
Hence, the correct answer is option: 'D'.