Let omega be a complex number such that 2omeg | vedclass

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

Question

(B) Given $2\omega + 1 = z$ and $z = \sqrt{-3} = i\sqrt{3}$.Thus,$\omega = \frac{i\sqrt{3} - 1}{2}$,which is a complex cube root of unity.We know that

Vedclass · Accepted Answer

$-z$

Vedclass · Answer

(B) Given $2\omega + 1 = z$ and $z = \sqrt{-3} = i\sqrt{3}$.Thus,$\omega = \frac{i\sqrt{3} - 1}{2}$,which is a complex cube root of unity.We know that $1 + \omega + \omega^2 = 0$,so $-\omega^2 - 1 = \omega$.Also,$\omega^3 = 1$,so $\omega^7 = \omega^{3 	imes 2 + 1} = \omega$.The determinant is $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{array} ight|$.Applying $R_1 	o R_1 + R_2 + R_3$:$\Delta = \left| \begin{array}{ccc} 3 & 1+\omega+\omega^2 & 1+\omega^2+\omega \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{array} ight| = \left| \begin{array}{ccc} 3 & 0 & 0 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{array} ight|$.Expanding along $R_1$: $\Delta = 3(\omega^2 - \omega^4) = 3(\omega^2 - \omega)$.Since $\omega = \frac{-1 + i\sqrt{3}}{2}$,$\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.$\Delta = 3\left( \frac{-1 - i\sqrt{3}}{2} - \frac{-1 + i\sqrt{3}}{2} ight) = 3\left( \frac{-2i\sqrt{3}}{2} ight) = -3i\sqrt{3} = -3z$.Given $\Delta = 3k$,we have $3k = -3z$,so $k = -z$.

Let $\omega$ be a complex number such that $2\omega + 1 = z$ where $z = \sqrt{-3}$. If $\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & -\omega^2 - 1 & \omega^2 \\ 1 & \omega^2 & \omega^7 \end{array} \right| = 3k$,then $k$ is equal to:

Similar Questions

The maximum value of $f(x) = \left|\begin{array}{ccc} \sin^{2} x & 1+\cos^{2} x & \cos 2x \\ 1+\sin^{2} x & \cos^{2} x & \cos 2x \\ \sin^{2} x & \cos^{2} x & \sin 2x \end{array}\right|, x \in R$ is:

Let $A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix}$ and $B$ be a matrix such that $B(I - A) = I + A$. Then the sum of the diagonal elements of $B^T B$ is equal to:

The total number of $3 \times 3$ matrices $A$ having entries from the set $\{0, 1, 2, 3\}$ such that the sum of all the diagonal entries of $AA^{T}$ is $9$,is equal to........

Let $A=\left[\begin{array}{ll}2 & 3 \\ a & 0\end{array}\right], a \in R$ be written as $P+Q$ where $P$ is a symmetric matrix and $Q$ is a skew-symmetric matrix. If $\operatorname{det}(Q)=9$,then the modulus of the sum of all possible values of the determinant of $P$ is equal to:

If $A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$ satisfies the equation $x^2 + 4x - p = 0$,then $p$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module