Let A denote the matrix left[begin{array}{ll} | vedclass

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

Question

(B) Given $A = \left[\begin{array}{ll}0 & i \ i & 0\end{array}ight]$.Calculating powers of $A$:$A^2 = \left[\begin{array}{ll}0 & i \ i & 0\end{arr

Vedclass · Accepted Answer

$\left[\begin{array}{ll}0 & i \ i & 0\end{array}ight]$

Vedclass · Answer

(B) Given $A = \left[\begin{array}{ll}0 & i \ i & 0\end{array}ight]$.Calculating powers of $A$:$A^2 = \left[\begin{array}{ll}0 & i \ i & 0\end{array}ight] \left[\begin{array}{ll}0 & i \ i & 0\end{array}ight] = \left[\begin{array}{cc}i^2 & 0 \ 0 & i^2\end{array}ight] = \left[\begin{array}{cc}-1 & 0 \ 0 & -1\end{array}ight] = -I$.$A^3 = A^2 \cdot A = (-I) \cdot A = -A$.$A^4 = A^2 \cdot A^2 = (-I) \cdot (-I) = I$.Since $A^4 = I$,the powers of $A$ repeat in a cycle of $4$: $I, A, -I, -A, I, \dots$.The sum of any four consecutive terms is $I + A + A^2 + A^3 = I + A - I - A = 0$.The series is $S = I + A + A^2 + \dots + A^{2010}$.There are $2011$ terms in total. Since $2011 = 4 	imes 502 + 3$,the sum consists of $502$ groups of $4$ terms (each summing to $0$) plus the remaining $3$ terms:$S = 502(0) + (I + A + A^2) = I + A - I = A$.Therefore,$S = \left[\begin{array}{ll}0 & i \ i & 0\end{array}ight]$.

Let $A$ denote the matrix $\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$,where $i^2=-1$,and let $I$ denote the identity matrix $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$. Then,$I+A+A^2+\ldots+A^{2010}$ is

Similar Questions

Let $A = [a_{ij}]$ be a square matrix of order $2$ with entries either $0$ or $1$. Let $E$ be the event that $A$ is an invertible matrix. Then the probability $P(E)$ is:

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P = [p_{ij}]$ be an $n \times n$ matrix with $p_{ij} = \omega^{i+j}$. Then $P^2 \neq 0$ when $n =$

$A=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \Rightarrow A^2-2A=$

Let $A^{-1} = \begin{bmatrix} 1 & 2017 & 2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8 \end{bmatrix}$. Then,$|2A| - |2A^{-1}|$ is equal to

Let $A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{bmatrix}$. The sum of the prime factors of $|P^{-1}AP - 2I|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module