KCET · Maths · Determinants
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
- A \( x y z \)
- B \( x^{-1} y^{-1} z^{-1} \)
- C \( -x-y-z \)
- D \( -1 \)
Answer & Solution
Correct Answer
(D) \( -1 \)
Step-by-step Solution
Detailed explanation
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
\[
\begin{array}{l}
\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1=0 \\
\Rightarrow x^{-1}+y^{-1}+z^{-1}=-1
\end{array}
\]
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