$\left| \begin{matrix} 2xy & x^{2} & y^{2} \ x^{2} & y^{2} & 2xy \ y^{2} & 2xy & x^{2} \end{matrix} ight|$ is equal to

Question

$\left| \begin{matrix} 2xy & x^{2} & y^{2} \ x^{2} & y^{2} & 2xy \ y^{2} & 2xy & x^{2} \end{matrix} ight|$ is equal to

Dr. Abubacker · Accepted Answer

(d)
$\left| \begin{matrix}
2xy & x^{2} & y^{2} \
x^{2} & y^{2} & 2xy \
y^{2} & 2xy & x^{2}
\end{matrix} ight| = 2xy\left( x^{2}y^{2} - 4x^{2}y^{2} ight) - x^{2}\left( x^{4} - 2xy^{3} ight) + y^{2}(2x^{3}y - y^{4})$
$= - 6x^{3}y^{3} - x^{6} + 2x^{3}y^{3} + 2x^{3}y^{3} - y^{6}$
$= - (x^{6} + y^{6} + 2x^{3}y^{3})$
$= - \left( x^{3} + y^{3} ight)^{2}$

Q. $\left| \begin{matrix} 2xy & x^{2} & y^{2} \\ x^{2} & y^{2} & 2xy \\ y^{2} & 2xy & x^{2} \end{matrix} \right|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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