Question

Let $\Delta=\begin{vmatrix}Ax & x ^2 & 1 \\ By & y ^2 & 1 \\ Cz & z ^2 & 1\end{vmatrix}$ and $\Delta_1=\begin{vmatrix}A & B & C \\ x & y & z \\ zy & zx & xy \end{vmatrix}$ then $\begin{vmatrix}A x & B y & C y \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1\end{vmatrix}$

• A. $\Delta_1 = - \Delta $
• B. $\Delta_1 = \Delta $
• C. $\Delta_1 \neq \Delta $
• D. $\Delta_1 = 2\, \Delta $

Answer 1

Answer: (B)

We have, $\Delta=\begin{bmatrix}A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{bmatrix}$ and
$\Delta_{1}=\begin{bmatrix}A & B & C \\x & y & z \\z y & z x & x y\end{bmatrix}$
Now, $ \Delta_{1}=\begin{bmatrix}A & B & C \\X & y & Z \\z y & z x & x y \end{bmatrix}$
On applying
$C_{1} \rightarrow x C_{1}, C_{2} \rightarrow y C_{2}, C_{3} \rightarrow z C_{3}$, we get
$=\frac{1}{x y z}\begin{bmatrix}A x & B y & C z \\x^{2} & y^{2} & z^{2} \\x y z & x y z & x y z \end{bmatrix}$
Taking common $xyz$ from $R_{3}$
$=\frac{x y z}{x y z}\begin{bmatrix}A x & B y & C z \\x^{2} & y^{2} & z^{2} \\1 & 1 & 1\end{bmatrix}$
$=\begin{bmatrix}A x & B y & C z \\x^{2} & y^{2} & z^{2} \\1 & 1 & 1\end{bmatrix}$
$=\begin{bmatrix}A x & x^{2} & 1 \\B y & y^{2} & 1 \\C z & z^{2} & 1\end{bmatrix}$
$|\because| A'| A |]$
$=\Delta$