(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

Let $S = \left\{ A = \begin{bmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{bmatrix} : a, b, c, d, e \in \{0, 1\} \text{ and } |A| \in \{-1, 1\} \right\}$,where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is:

MediumIIT 2024

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MediumIIT 1984

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EasyGUJCET 2017

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Evaluate $\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|$

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If $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 3 \\ 0 & 0 & 2 \end{bmatrix}$,then $|AB|$ is equal to

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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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