If A=left[begin{array}{rr}i & -i -i & iend{a | vedclass

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(D) Given,$A=\left[\begin{array}{rr}i & -i \ -i & i\end{array}ight]$ and $B=\left[\begin{array}{rr}1 & -1 \ -1 & 1\end{array}ight]$.First,calcul

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$128$

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(D) Given,$A=\left[\begin{array}{rr}i & -i \ -i & i\end{array}ight]$ and $B=\left[\begin{array}{rr}1 & -1 \ -1 & 1\end{array}ight]$.First,calculate $A^2$:$A^2 = \left[\begin{array}{rr}i & -i \ -i & i\end{array}ight] \left[\begin{array}{rr}i & -i \ -i & i\end{array}ight] = \left[\begin{array}{rr}i^2+i^2 & -i^2-i^2 \ -i^2-i^2 & i^2+i^2\end{array}ight] = \left[\begin{array}{rr}-2 & 2 \ 2 & -2\end{array}ight] = -2B$.Next,calculate $B^2$:$B^2 = \left[\begin{array}{rr}1 & -1 \ -1 & 1\end{array}ight] \left[\begin{array}{rr}1 & -1 \ -1 & 1\end{array}ight] = \left[\begin{array}{rr}2 & -2 \ -2 & 2\end{array}ight] = 2B$.Now,calculate $A^8$:$A^8 = (A^2)^4 = (-2B)^4 = (-2)^4 B^4 = 16 B^4$.Since $B^4 = (B^2)^2 = (2B)^2 = 4B^2 = 4(2B) = 8B$.Therefore,$A^8 = 16(8B) = 128B$.

If $A=\left[\begin{array}{rr}i & -i \\ -i & i\end{array}\right]$ and $B=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right]$,then find $A^8$. (in $B$)

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