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?

The characteristic polynomial of $A$ is
\begin{align}
        \det(A-\lambda I)=\left|\begin{array}{c}
                                  \frac{5}{2}-\lambda & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2}-\lambda
                                \end{array}\right|\\
          = (5/2-\lambda)(-1/2-\lambda)+9/4 \\
          = -5/4-2\lambda+\lambda^2+9/4 \\
          = (\lambda-1)^2
\end{align}
Cayley-Hamilton gives $(A-I)^2=0$. 
So, a binomial expansion works great because only two terms are non-zero:
\begin{align}
     A^{2022} &= (I+(A-I))^{2022} \\
              &=I+2022 (A-I) \\
              &=2022 A - 2021 I \\
              &=2022\left[\begin{array}{rr}
                           \frac{5}{2} & \frac{3}{2} \\
                           -\frac{3}{2} & -\frac{1}{2}
                          \end{array}\right]
              -\left[\begin{array}{rr}2021 & 0 \\ 0 & 2021\end{array}\right] \\
              &=\left[\begin{array}{rr}
                      3034 & 3033 \\
                     -3033 & -3032
                     \end{array}\right].
\end{align}
