Let z = frac{-1 + sqrt{3}i}{2}, where i = sqr | vedclass

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

Question

(C) Given $z = \frac{-1}{2} + i \frac{\sqrt{3}}{2} = \omega$, where $\omega$ is the cube root of unity. Thus, $\omega^3 = 1$ and $1 + \omega + \omega^

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given $z = \frac{-1}{2} + i \frac{\sqrt{3}}{2} = \omega$, where $\omega$ is the cube root of unity. Thus, $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.$P = \begin{bmatrix} (-1)^r \omega^r & \omega^{2s} \ \omega^{2s} & \omega^r \end{bmatrix}$.$P^2 = \begin{bmatrix} (-1)^r \omega^r & \omega^{2s} \ \omega^{2s} & \omega^r \end{bmatrix} \begin{bmatrix} (-1)^r \omega^r & \omega^{2s} \ \omega^{2s} & \omega^r \end{bmatrix} = \begin{bmatrix} (-1)^{2r} \omega^{2r} + \omega^{4s} & (-1)^r \omega^{r+2s} + \omega^{r+2s} \ (-1)^r \omega^{r+2s} + \omega^{r+2s} & \omega^{4s} + \omega^{2r} \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$.For the off-diagonal elements to be zero: $((-1)^r + 1) \omega^{r+2s} = 0$. Since $\omega 
eq 0$, we must have $(-1)^r + 1 = 0$, which implies $r$ must be odd. Thus, $r \in \{1, 3\}$.Case $1$: $r = 1$. The diagonal elements give $\omega^2 + \omega^{4s} = -1$. Since $1 + \omega + \omega^2 = 0$, we have $\omega^2 + \omega = -1$. Thus, $\omega^{4s} = \omega$, which means $4s \equiv 1 \pmod 3$, so $s \equiv 1 \pmod 3$. For $s \in \{1, 2, 3\}$, $s = 1$.Case $2$: $r = 3$. The diagonal elements give $\omega^6 + \omega^{4s} = -1$. Since $\omega^3 = 1$, $1 + \omega^{4s} = -1$, so $\omega^{4s} = -2$. This is impossible as $|\omega^{4s}| = 1$.Thus, the only valid pair is $(r, s) = (1, 1)$. The total number of pairs is $1$.

Let $z = \frac{-1 + \sqrt{3}i}{2}$, where $i = \sqrt{-1}$, and $r, s \in \{1, 2, 3\}$. Let $P = \begin{bmatrix} (-z)^r & z^{2s} \\ z^{2s} & z^r \end{bmatrix}$ and $I$ be the identity matrix of order $2$. Then the total number of ordered pairs $(r, s)$ for which $P^2 = -I$ is

Similar Questions

If $P$ is a non-singular matrix such that $I+P+P^2+\ldots+P^{n}=0$ ($0$ denotes the null matrix), then $P^{-1}=$

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

Let $\alpha, \beta$ be the roots of the equation $ax^2+bx+c=0$, where $a, b, c$ are real. If $s_n = \alpha^n + \beta^n$ and $\left|\begin{array}{ccc}3 & 1+s_1 & 1+s_2 \\ 1+s_1 & 1+s_2 & 1+s_3 \\ 1+s_2 & 1+s_3 & 1+s_4\end{array}\right| = k \frac{(a+b+c)^2}{a^4}$, then $k =$

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}$,then show that $A^{3} - 23A - 40I = 0$.

$A$ and $B$ are two non-singular square matrices of order $3 \times 3$ such that $AB = A$ and $|A + B| \neq 0$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module