If alpha = cos frac{pi}{3} + i sin frac{pi}{3 | vedclass

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

Question

(D) Given $\alpha = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = e^{i \pi / 3}$.Note that $\alpha^3 = e^{i \pi} = -1$ and $\alpha^6 = 1$.The determinan

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given $\alpha = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = e^{i \pi / 3}$.Note that $\alpha^3 = e^{i \pi} = -1$ and $\alpha^6 = 1$.The determinant is $D = \left| \begin{array}{ccc} 1 & \alpha & \alpha^2 \ \alpha^2 & 1 & \alpha \ \alpha & \alpha^2 & 1 \end{array} ight|$.Applying the operation $C_1 	o C_1 + C_2 + C_3$,we get:$D = \left| \begin{array}{ccc} 1 + \alpha + \alpha^2 & \alpha & \alpha^2 \ 1 + \alpha + \alpha^2 & 1 & \alpha \ 1 + \alpha + \alpha^2 & \alpha^2 & 1 \end{array} ight| = (1 + \alpha + \alpha^2) \left| \begin{array}{ccc} 1 & \alpha & \alpha^2 \ 1 & 1 & \alpha \ 1 & \alpha^2 & 1 \end{array} ight|$.Expanding the determinant:$D = (1 + \alpha + \alpha^2) [1(1 - \alpha^3) - \alpha(1 - \alpha) + \alpha^2(\alpha^2 - 1)]$.Since $\alpha^3 = -1$,$1 - \alpha^3 = 2$.$D = (1 + \alpha + \alpha^2) [2 - \alpha + \alpha^2 + \alpha^4 - \alpha^2] = (1 + \alpha + \alpha^2) [2 - \alpha + \alpha^4]$.Since $\alpha = \frac{1}{2} + i \frac{\sqrt{3}}{2}$,$\alpha^2 = \alpha - 1$ is not true here,but $1 + \alpha + \alpha^2 = \frac{1 - \alpha^3}{1 - \alpha} = \frac{2}{1 - \alpha}$.Alternatively,calculating directly: $D = 1(1 - \alpha^3) - \alpha(\alpha^2 - \alpha^2) + \alpha^2(\alpha^4 - \alpha) = 1(1 - (-1)) - 0 + \alpha^2(\alpha^4 - \alpha) = 2 + \alpha^6 - \alpha^3 = 2 + 1 - (-1) = 4$.

If $\alpha = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}$,then the value of the determinant $\left| \begin{array}{ccc} 1 & \alpha & \alpha^2 \\ \alpha^2 & 1 & \alpha \\ \alpha & \alpha^2 & 1 \end{array} \right|$ is

Similar Questions

The number of integers $x$ satisfying $-3 x^4 + \operatorname{det}\begin{bmatrix} 1 & x & x^2 \\ 1 & x^2 & x^4 \\ 1 & x^3 & x^6 \end{bmatrix} = 0$ is equal to

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

If $A=\left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$,then $A+A^3+A^4+A^5+3 I=$

Let $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$. Let $\alpha, \beta \in \mathbb{R}$ be such that $\alpha A^{2} + \beta A = 2I$. Then $\alpha + \beta$ is equal to -

Let $A$ be a square matrix such that $AA^T = I$. Then $\frac{1}{2} A[(A+A^T)^2 + (A-A^T)^2]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module