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(B) Given $A = \begin{bmatrix} 1 & 1 \ 0 & i \end{bmatrix}$.Calculate the powers of $A$:$A^2 = \begin{bmatrix} 1 & 1 \ 0 & i \end{bmatrix} \begin{bm

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(B) Given $A = \begin{bmatrix} 1 & 1 \ 0 & i \end{bmatrix}$.Calculate the powers of $A$:$A^2 = \begin{bmatrix} 1 & 1 \ 0 & i \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & i \end{bmatrix} = \begin{bmatrix} 1 & 1+i \ 0 & i^2 \end{bmatrix} = \begin{bmatrix} 1 & 1+i \ 0 & -1 \end{bmatrix}$.$A^3 = A^2 	imes A = \begin{bmatrix} 1 & 1+i \ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & i \end{bmatrix} = \begin{bmatrix} 1 & 1+i+i \ 0 & -i \end{bmatrix} = \begin{bmatrix} 1 & 1+2i \ 0 & -i \end{bmatrix}$.Wait, let us re-evaluate $A^4$:$A^4 = A^2 	imes A^2 = \begin{bmatrix} 1 & 1+i \ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1+i \ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 1+i-1-i \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.Since $A^4 = I$, we have $A^{2018} = A^{2016} 	imes A^2 = (A^4)^{504} 	imes A^2 = I^{504} 	imes A^2 = A^2$.Thus, $A^{2018} = \begin{bmatrix} 1 & 1+i \ 0 & -1 \end{bmatrix}$.Comparing with $\begin{bmatrix} a & b \ c & d \end{bmatrix}$, we get $a=1$ and $d=-1$.Therefore, $a+d = 1 + (-1) = 0$.

If $A=\begin{bmatrix} 1 & 1 \\ 0 & i \end{bmatrix}$ and $A^{2018}=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then $(a+d)$ equals

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