Let omega be a complex cube root of unity wit | vedclass

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

Question

(A) Given $P = \begin{bmatrix} \omega^{1+1} & \omega^{1+2} \ \omega^{2+1} & \omega^{2+2} \end{bmatrix} = \begin{bmatrix} \omega^2 & \omega^3 \ \omeg

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(A) Given $P = \begin{bmatrix} \omega^{1+1} & \omega^{1+2} \ \omega^{2+1} & \omega^{2+2} \end{bmatrix} = \begin{bmatrix} \omega^2 & \omega^3 \ \omega^3 & \omega^4 \end{bmatrix} = \begin{bmatrix} \omega^2 & 1 \ 1 & \omega \end{bmatrix}$.Calculate $P^2 = \begin{bmatrix} \omega^2 & 1 \ 1 & \omega \end{bmatrix} \begin{bmatrix} \omega^2 & 1 \ 1 & \omega \end{bmatrix} = \begin{bmatrix} \omega^4+1 & \omega^2+\omega \ \omega^2+\omega & 1+\omega^2 \end{bmatrix} = \begin{bmatrix} \omega+1 & -1 \ -1 & -\omega \end{bmatrix} = \begin{bmatrix} -\omega^2 & -1 \ -1 & -\omega \end{bmatrix} = -P$.Since $P^2 = -P$,we have $P^3 = P^2 \cdot P = (-P) \cdot P = -P^2 = -(-P) = P$.Thus,$P^3 = P$. For $P^k = P$,$k$ must be an odd integer greater than or equal to $3$. Since $P^3 = P$,$P^5 = P^3 \cdot P^2 = P \cdot (-P) = -P^2 = P$. In general,$P^k = P$ for any odd $k \geq 3$.Checking the options,$k$ must be odd. Among the given options,$57$ is the only odd number.

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P = [p_{ij}]$ be a $2 \times 2$ matrix with $p_{ij} = \omega^{i+j}$. For $P^2 \neq 0$,if $P^k = P$,then $k$ is equal to

Similar Questions

$\left|\begin{array}{ll}2 & 1 \\ 3 & 1\end{array}\right|+\left|\begin{array}{cc}1 & 1/3 \\ 3 & 1\end{array}\right|+\left|\begin{array}{cc}1/2 & 1/9 \\ 3 & 1\end{array}\right|+\left|\begin{array}{cc}1/4 & 1/27 \\ 3 & 1\end{array}\right|+\ldots \infty=$

If $\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ is a skew-symmetric matrix and $b, c, f$ are non-zero real numbers, then $\frac{b}{c} = $

If $\left| \begin{array}{ccc} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{array} \right| = 0$ and the vectors $\vec{a} = (1, a, a^2)$,$\vec{b} = (1, b, b^2)$,and $\vec{c} = (1, c, c^2)$ are non-coplanar,then $abc$ is equal to

If $z = \begin{bmatrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{bmatrix}$, then which of the following is true? (where $i = \sqrt{-1}$)

Let $A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],$ show that $(a \mathrm{I}+b \mathrm{A})^{n}=a^{n} \mathrm{I}+n a^{n-1} b \mathrm{A},$ where $\mathrm{I}$ is the identity matrix of order $2$ and $n \in \mathrm{N}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module