left| begin{array}{ccc} 1 & 1 & 1 1 & omega^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $D = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & \omega^2 & \omega \ 1 & \omega & \omega^2 \end{array} ight|$.Applying $C_1 	o C_1 - C_2$ a

Vedclass · Accepted Answer

$3\sqrt{3}i$

Vedclass · Answer

(A) Let $D = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & \omega^2 & \omega \ 1 & \omega & \omega^2 \end{array} ight|$.Applying $C_1 	o C_1 - C_2$ and $C_2 	o C_2 - C_3$:$D = \left| \begin{array}{ccc} 0 & 0 & 1 \ 1-\omega^2 & \omega^2-\omega & \omega \ 1-\omega & \omega-\omega^2 & \omega^2 \end{array} ight|$.Expanding along the first row:$D = 1 \cdot [(1-\omega^2)(\omega-\omega^2) - (1-\omega)(\omega^2-\omega)]$$D = (\omega - \omega^2 - \omega^2 + \omega^4) - (\omega^2 - \omega - \omega^3 + \omega^2)$Since $\omega^3 = 1$ and $\omega^4 = \omega$:$D = (\omega - 2\omega^2 + \omega) - (2\omega^2 - \omega - 1)$$D = 2\omega - 2\omega^2 - 2\omega^2 + \omega + 1 = 3\omega - 4\omega^2 + 1$.Using $1 + \omega + \omega^2 = 0$,we have $1 = -\omega - \omega^2$:$D = 3\omega - 4\omega^2 - \omega - \omega^2 = 2\omega - 5\omega^2$.Alternatively,calculating directly:$D = 1(\omega^4 - \omega^2) - 1(\omega^2 - \omega) + 1(\omega - \omega^2) = \omega - \omega^2 - \omega^2 + \omega + \omega - \omega^2 = 3\omega - 3\omega^2 = 3(\omega - \omega^2)$.Substituting $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$:$D = 3 \left( \frac{-1 + i\sqrt{3}}{2} - \frac{-1 - i\sqrt{3}}{2} ight) = 3 \left( \frac{2i\sqrt{3}}{2} ight) = 3i\sqrt{3}$.

$\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{array} \right| = $

Similar Questions

Find the area of the triangle with vertices $(a \cos \theta, b \sin \theta)$,$(-a \sin \theta, b \cos \theta)$,and $(-a \cos \theta, -b \sin \theta)$.

Let $A = \begin{bmatrix} -2 & x & 1 \\ x & 1 & 1 \\ 2 & 3 & -1 \end{bmatrix}$. If the roots of the equation $\operatorname{det}(A) = 0$ are $l$ and $m$,then find the value of $l^3 - m^3$.

The set of all values of $\lambda$ for which the system of linear equations $2x_1 - 2x_2 + x_3 = \lambda x_1$,$2x_1 - 3x_2 + 2x_3 = \lambda x_2$,and $-x_1 + 2x_2 = \lambda x_3$ has a non-trivial solution:

If the points with position vectors $60 \hat{i}+3 \hat{j}$,$40 \hat{i}-8 \hat{j}$,and $a \hat{i}-52 \hat{j}$ are collinear,then $a$ is equal to

If $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module