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

Question

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, $ \Delta_{1}=-\Delta $, $ \Delta_{1}=\Delta $, $ \Delta_{1} \neq \Delta $, $ \Delta_{1}=2 \Delta $

MARKS App · Accepted Answer

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} $

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