Question

If $△=\left|\begin{array}{ccc}ax& x^2& 1\\ by& y^2& 1\\ cz& z^2& 1\end{array}\right|$ and
${△}_1=\left|\begin{array}{ccc}a& b& c\\ x& y& z\\ yz& zx& xy\end{array}\right|,$ then which of the following is correct

• A. ${△}_1+△=0$
• B. ${△}_1\ne △$
• C. ${△}_1-△=0$
• D. ${△}_1-2△=0$

Answer 1

The correct option is C ${△}_1-△=0$

Given : ${△}_1=\left|\begin{array}{ccc}a& b& c\\ x& y& z\\ yz& zx& xy\end{array}\right|$
Taking transpose of above determinant, we get
$\Rightarrow {△}_1=\left|\begin{array}{ccc}a& x& zy\\ b& y& zx\\ c& z& xy\end{array}\right|$
Now multiplying $R_1$ by $x$, $R_2$ by $y$ and $R_3,$ by $z$,we get
$\Rightarrow {△}_1=\frac{1}{xyz}\left|\begin{array}{ccc}ax& x^2& xzy\\ by& y^2& xyz\\ cz& z^2& xyz\end{array}\right|$
Taking $xyz$ common from $C_3,$ we get
$\Rightarrow {△}_1=\frac{xyz}{xyz}\left|\begin{array}{ccc}ax& x^2& 1\\ by& y^2& 1\\ cz& z^2& 1\end{array}\right|=△$
$\Rightarrow {△}_1-△=0$