<s>[INST] Let $A= \begin{pmatrix} 8&2 \\ -8&-2 \end{pmatrix}$. Find the entry in the first row and second column of $A^{2014}$ I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.
Where:
$P=\begin{pmatrix}  1&1 \\   -4&-1  \end{pmatrix}$
$D=\begin{pmatrix}  0&0 \\   0&6  \end{pmatrix}$
$P^{-1}=\frac{1}{3}\begin{pmatrix}  -1&-1 \\   4&1  \end{pmatrix}$.  
So that $A^{2014}=PD^{2014}P^{-1}$
But i just want to know whether there is an alternate method.  
 [/INST] Hint Show that
$$A^2=6A$$
Note: You can show the stronger statement:
$$\begin{pmatrix} 8&2 \\ -8&-2 \end{pmatrix}\begin{pmatrix} a&b \\ -a&-b \end{pmatrix}=6\begin{pmatrix} a&b \\ -a&-b \end{pmatrix}$$
but this is overkill.
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