Q86 / 180 · KCET 2017 · Mathematics
Let \( \Delta=\left|\begin{array}{ccc}A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1\end{array}\right| \) and
\( \Delta_{1}=\left|\begin{array}{cccc}A & B & C & \\ x & y & z & \\ z y & z x & x y & \end{array}\right| \) then
- A\( \Delta_{1}=-\Delta \)
- Correct answerB\( \Delta_{1}=\Delta \)
- C\( \Delta_{1} \neq \Delta \)
- D\( \Delta_{1}=2 \Delta \)
Solution
Given that, \( \Delta=\left|\begin{array}{ccc}A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1\end{array}\right| \rightarrow(1) \)
\[ \text { and } \Delta_{1}=\left|\begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y \end{array}\right| \rightarrow(2) \]
Take out common \( x, y \) and \( z \) in Eq. (1) from \( R_{1}, R_{2} \) and \( R_{3} \) respectively we get
\( \Delta=x y z\left|\begin{array}{ccc}A & x & \frac{1}{x} \\ B & y & \frac{1}{y} \\ C & z & \frac{1}{z}\end{array}\right| \)
\( C_{3} \rightarrow x y z C_{3} \), we get
\( \Delta=\left|\begin{array}{ccc}A & x & y z \\ B & y & z x \\ C & z & x y\end{array}\right| \)
\( =\left|\begin{array}{ccc}A & B & C \\ x & y & z \\ y z & z x & x y\end{array}\right|=\Delta_{1} \)
Therefore, \( \Delta=\Delta_{1} \)
\[ \text { and } \Delta_{1}=\left|\begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y \end{array}\right| \rightarrow(2) \]
Take out common \( x, y \) and \( z \) in Eq. (1) from \( R_{1}, R_{2} \) and \( R_{3} \) respectively we get
\( \Delta=x y z\left|\begin{array}{ccc}A & x & \frac{1}{x} \\ B & y & \frac{1}{y} \\ C & z & \frac{1}{z}\end{array}\right| \)
\( C_{3} \rightarrow x y z C_{3} \), we get
\( \Delta=\left|\begin{array}{ccc}A & x & y z \\ B & y & z x \\ C & z & x y\end{array}\right| \)
\( =\left|\begin{array}{ccc}A & B & C \\ x & y & z \\ y z & z x & x y\end{array}\right|=\Delta_{1} \)
Therefore, \( \Delta=\Delta_{1} \)
Hence, the correct answer is option: 'B'.