Diagonalizing a 2×2 matrix

Eigenvalues and eigenvectors. From the previous exercise, the eigenvalues are $2$ and $5$, with eigenvectors $v_1 = (1,-2)$ and $v_2 = (1,1)$.

Matrices $P$ and $D$.
$$P = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}, \qquad D = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}.$$

Inverse of $P$. $\det P = 1\cdot1 - 1\cdot(-2) = 3$, so
$$P^{-1} = \frac13\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}.$$

Conclusion. $A = PDP^{-1}$ with the matrices above. Consequently,
$$A^k = PD^kP^{-1} = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}\begin{pmatrix} 2^k & 0 \\ 0 & 5^k \end{pmatrix}\frac13\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}.$$