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

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(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

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

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(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

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