(B) Given the determinant $\Delta = \begin{vmatrix} 1 & 2\omega \\ \omega & \omega^2 \end{vmatrix}$.Expanding the determinant,we get $\Delta = (1)(\om

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(B) Given the determinant $\Delta = \begin{vmatrix} 1 & 2\omega \\ \omega & \omega^2 \end{vmatrix}$.Expanding the determinant,we get $\Delta = (1)(\omega^2) - (2\omega)(\omega) = \omega^2 - 2\omega^2 = -\omega^2$.Now,we need to find $\Delta^2$.$\Delta^2 = (-\omega^2)^2 = \omega^4$.Since $\omega$ is a cube root of unity,we know that $\omega^3 = 1$.Therefore,$\Delta^2 = \omega^3 \times \omega = 1 \times \omega = \omega$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $\omega$ is a cube root of unity and $\Delta = \begin{vmatrix} 1 & 2\omega \\ \omega & \omega^2 \end{vmatrix}$,then $\Delta^2$ is equal to

A
$-\omega$
B
$\omega$
C
$1$
D
$\omega^2$

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3-4. Matrices and Determinants

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $\omega$ is a cube root of unity and $\Delta = \begin{vmatrix} 1 & 2\omega \\ \omega & \omega^2 \end{vmatrix}$,then $\Delta^2$ is equal to

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