Answer
$A^n=\left[\begin{array}{c c} 2^{n-1}& 2^{n-1}\\ 2^{n-1}& 2^{n-1} \end{array} \right]$
Work Step by Step
We detect the pattern as follows:
$A^2=\left[\begin{array}{c c} 2& 2\\ 2& 2\end{array} \right], A^3=\left[\begin{array}{c c} 4&4\\ 4& 4\end{array} \right],A^4=\left[\begin{array}{c c} 8& 8\\ 8& 8\end{array} \right]$
Hence, according to the pattern
$A^n=\left[\begin{array}{c c} 2^{n-1}& 2^{n-1}\\ 2^{n-1}& 2^{n-1} \end{array} \right]$
