(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 and 4 .Determinants and Matrices 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 and 4 .Determinants and Matrices

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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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DifficultJEE MAIN 2020

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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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If $\left|\begin{array}{ccc}0 & ab^2 & ac^2 \\ a^2b & 0 & bc^2 \\ a^2c & b^2c & 0\end{array}\right|=m(abc)^k$,then $m+k=$ . . . . . . .

If the system of equations $ax + y + z = 0$,$x + by + z = 0$ and $x + y + cz = 0$,where $a, b, c \neq 1$,has a non-trivial solution,then the value of $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$ is

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