Using factorisation, $A=PJP^{-1}$ to compute $A^k$ Using factorisation, $A=PJP^{-1}$ to compute $A^k$, where $k$ represents an arbitrary positive integer.
$$ \begin{bmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{-1} & \mathbf{2} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{1} \\ \mathbf{0} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{-1} & \mathbf{1} \end{bmatrix}. $$
Not sure how to finish this problem as the matrix $J$ is not diagonal, but $J^k$ can still be found somehow. How exactly is $J^k$ obtained in this case?

You can easily prove (for example with induction) that 
$${\begin{pmatrix} 1 & 1  \\ 0 & 1  \\  \end{pmatrix}}^k=\begin{pmatrix} 1 & k  \\ 0 & 1  \\  \end{pmatrix}$$
Now $A^k$ can be computed as follows:
$A^k=(S^{-1}BS)^{k}=\underbrace{S^{-1}BS\cdot S^{-1}BS\dots   S^{-1}BS}_{k \text{ times }}=S^{-1}B^{k}S=S^{-1}\begin{pmatrix} 1 & k  \\ 0 & 1  \\  \end{pmatrix}S$
