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If $A=\left(\begin{array}{ll}1 & 1 \\ 0 & i\end{array}\right)$ and $A^{2018}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$, then $(a+d)$ equals
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$A^2=\left(\begin{array}{cc}1 & 1+i \\ 0 & -1\end{array}\right), A^3=\left(\begin{array}{cc}1 & i \\ 0 & -i\end{array}\right), A^4=I$
$\begin{aligned}
&\therefore A^{2018}=A^{2016} \times A^2=A^2=\left(\begin{array}{ll}
1 & 1+i \\
0 & -1
\end{array}\right) \\
&\therefore a+d=1+(-1)=0
\end{aligned}$
$\begin{aligned}
&\therefore A^{2018}=A^{2016} \times A^2=A^2=\left(\begin{array}{ll}
1 & 1+i \\
0 & -1
\end{array}\right) \\
&\therefore a+d=1+(-1)=0
\end{aligned}$
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