Let omega be the complex number cos frac{2 pi | vedclass

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

Question

(A) Given $\omega = e^{i 2 \pi / 3}$,we know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.Let the determinant be $\Delta = \left|\begin{array}{

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given $\omega = e^{i 2 \pi / 3}$,we know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.Let the determinant be $\Delta = \left|\begin{array}{ccc} z+1 & \omega & \omega^2 \ \omega & z+\omega^2 & 1 \ \omega^2 & 1 & z+\omega \end{array}ight|$.Applying the operation $C_1 	o C_1 + C_2 + C_3$,we get:$\Delta = \left|\begin{array}{ccc} z+1+\omega+\omega^2 & \omega & \omega^2 \ z+\omega+\omega^2+1 & z+\omega^2 & 1 \ z+\omega^2+1+\omega & 1 & z+\omega \end{array}ight| = \left|\begin{array}{ccc} z & \omega & \omega^2 \ z & z+\omega^2 & 1 \ z & 1 & z+\omega \end{array}ight|$.Factoring out $z$ from the first column:$\Delta = z \left|\begin{array}{ccc} 1 & \omega & \omega^2 \ 1 & z+\omega^2 & 1 \ 1 & 1 & z+\omega \end{array}ight|$.Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\Delta = z \left|\begin{array}{ccc} 1 & \omega & \omega^2 \ 0 & z+\omega^2-\omega & 1-\omega^2 \ 0 & 1-\omega & z+\omega-\omega^2 \end{array}ight| = z[(z+\omega^2-\omega)(z+\omega-\omega^2) - (1-\omega^2)(1-\omega)]$.Expanding this,we find $\Delta = z(z^2) = z^3 = 0$.Thus,$z = 0$ is the only solution. The number of distinct complex numbers is $1$.

Let $\omega$ be the complex number $\cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3}$. Then the number of distinct complex numbers $z$ satisfying $\left|\begin{array}{ccc} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{array}\right| = 0$ is equal to

Similar Questions

Let $A$ and $B$ be real matrices of the form $\begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}$ and $\begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix}$,respectively.
Statement $1$: $AB - BA$ is always an invertible matrix.
Statement $2$: $AB - BA$ is never an identity matrix.

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ a & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}$ and $|A| = 2$. If $|2 \operatorname{adj}(2 \operatorname{adj}(2 A))| = 32^n$,then $3n + \alpha$ is equal to:

Let $B=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$ and $A$ be a $2 \times 2$ matrix satisfying $\left(A^T\right)^{-1}=A$. If $X=A B A^T$, then $A^T X^{2021} A=$

If $P$ and $Q$ are square matrices such that $P^{2006} = O$ and $PQ = P + Q$,then $\det(Q)$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module