Question

If ${\Delta }_1=\left|\begin{array}{cc}\omega & -{\omega }^2\\ -{\omega }^2& -\omega \end{array}\right|,{\Delta }_2=\left|\begin{array}{cc}-{\omega }^2& -\omega \\ \omega & {\omega }^2\end{array}\right|.$ Then the value of ${\Delta }_1\cdot {\Delta }_2=$
(where $\omega$ is a cube root of unity)

• A. $1+\omega$
• B. $-{\omega }^2+\omega$
• C. $-1-\omega$
• D. $-\omega +{\omega }^2$

Answer 1

The correct option is D $-\omega +{\omega }^2$

${\Delta }_1\cdot {\Delta }_2=\left|\begin{array}{cc}\omega & -{\omega }^2\\ -{\omega }^2& -\omega \end{array}\right|\cdot \left|\begin{array}{cc}-{\omega }^2& -\omega \\ \omega & {\omega }^2\end{array}\right|=\left|\begin{array}{cc}\omega (-{\omega }^2)-{\omega }^2\left(\omega \right)& \omega (-\omega )-{\omega }^2\left({\omega }^2\right)\\ -{\omega }^2(-{\omega }^2)-\omega \left(\omega \right)& -{\omega }^2(-\omega )-\omega \left({\omega }^2\right)\end{array}\right|=\left|\begin{array}{cc}-2& -{\omega }^2-\omega \\ \omega -{\omega }^2& 0\end{array}\right|$
$\Rightarrow {\Delta }_1\cdot {\Delta }_2=0-(-{\omega }^2+{\omega }^4)=-(-{\omega }^2+\omega )$
[$\because {\omega }^3=1$]
$\therefore {\Delta }_1\cdot {\Delta }_2={\omega }^2-\omega$