If the matrix A is given, then find the value of $A^{2022}$ If it is given that matrix $ A = \begin{bmatrix}
    \dfrac{5}{2} & \dfrac{3}{2}\\
    -\dfrac{3}{2} & -\dfrac{1}{2} 
\end{bmatrix}$ then find the value of $A^{2022}$.
Here is my try on it : $$\\$$
If we look at the pattern while multliplying A with itself it comes out to be something like this,
$$A^2 = \begin{bmatrix}
    4 & 3\\
    -3 & -2
\end{bmatrix}$$
$$ A^3 = \begin{bmatrix}
    \dfrac{11}{2} & \dfrac{9}{2}\\
    -\dfrac{9}{2} & -\dfrac{7}{2} 
\end{bmatrix}$$
$$A^4 = \begin{bmatrix}
    7 & 6\\
    -6 & -5
\end{bmatrix}$$
so $$ A^n = \begin{bmatrix}
    \dfrac{3n}{2}+1 & \dfrac{3n}{2}\\
    -\dfrac{3n}{2} & -\dfrac{3n}{2}+1 
\end{bmatrix} \tag{1}\label{eq1}$$
thus after substituting $n=2022$  we get $ A^{2022} = \begin{bmatrix}
    3034 & 3033\\
    -3033 & -3032
\end{bmatrix}$
$$\\$$
Is there is any other way to find the value of $A^{2022}$ without finding out the pattern and then general equation \eqref{eq1} like I did above?

COMMENT.-$A$ has a double eigenvalue $\lambda=1$, is not diagonalizable and its Jordan matrix decomposition is $$A=\begin{pmatrix}-1&-\frac23\\1&0\end{pmatrix}*\begin{pmatrix}1&1\\0&1\end{pmatrix}*\begin{pmatrix}0&1\\-\frac32&-\frac32\end{pmatrix}^{-1}$$ so we have
$$A^{2022}=\begin{pmatrix}-1&-\frac23\\1&0\end{pmatrix}*\begin{pmatrix}1&1\\0&1\end{pmatrix}^{2022}*\begin{pmatrix}0&1\\-\frac32&-\frac32\end{pmatrix}^{-1}$$
