If $A =\begin{bmatrix}i&-i\\-i&i\end{bmatrix}$ and $B =\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$, then $A^8$ equals

128B –128B 4B –64B

128B

$A^2 = B^2 ⇒ A^2 =\begin{vmatrix}2&-2\\-2&2\end{vmatrix}= –2B$ $A^4 = (–2B)^2 = 4B^2 = 4(2B) = 8B$ $A^8 = 64B^2 = 128B$