If A=(2102)A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}A=(2012), find AnA^nAn.
(2nn2n−102n)\begin{pmatrix} 2^n & n 2^{n-1} \\ 0 & 2^n \end{pmatrix}(2n0n2n−12n)
(2nn2n02n)\begin{pmatrix} 2^n & n 2^n \\ 0 & 2^n \end{pmatrix}(2n0n2n2n)
(2n002n)\begin{pmatrix} 2^n & 0 \\ 0 & 2^n \end{pmatrix}(2n002n)
(n2nn0n2n)\begin{pmatrix} n 2^n & n \\ 0 & n 2^n \end{pmatrix}(n2n0nn2n)