Question

$\Delta_{1}=\begin{vmatrix}y^{5} z^{6}\left(z^{3}-y^{3}\right) & x^{4} z^{6}\left(x^{3}-z^{3}\right) & x^{4} y^{5}\left(y^{3}-x^{3}\right) \\ y^{2} z^{3}\left(y^{6}-z^{6}\right) & x z^{3}\left(z^{6}-x^{6}\right) & x y^{2}\left(x^{6}-y^{6}\right) \\ y^{2} z^{3}\left(z^{3}-y^{3}\right) & x z^{3}\left(x^{3}-z^{3}\right) & x y^{2}\left(y^{3}-x^{3}\right)\end{vmatrix}$ and
$\Delta_{2}=\begin{vmatrix}x & y^{2} & z^{3} \\ x^{4} & y^{5} & z^{6} \\ x^{7} & y^{8} & z^{9}\end{vmatrix}$ Then $\Delta_{1} \Delta_{2}$ is equal to

• A. $\Delta_{2}^{3}$
• B. $\Delta_{2}^{2}$
• C. $\Delta_{2}^{4}$
• D. None of these

Answer 1

Answer: (A)

The given determinant $\Delta_{1}$ is obtained by corresponding cofactors of determinant $\Delta_{2}$; hence $\Delta_{1}=\Delta_{2}^{2}$.
Now $\Delta_{ L } \Delta_{2}=\Delta_{2}^{2} \Delta_{2}=\Delta_{2}^{3}$