If $$M = \begin{pmatrix} \dfrac{5}{2} & \dfrac{3}{2} \\ -\dfrac{3}{2} & -\dfrac{1}{2} \end{pmatrix}$$, then which of the following matrices is equal to $$M^{2022}$$?

Write the given matrix in the form $$M = I + N$$, where $$I$$ is the identity matrix and $$N$$ is easy to handle.

First find $$I$$ minus $$M$$:
$$M = \begin{pmatrix}\dfrac52 & \dfrac32 \\[4pt] -\dfrac32 & -\dfrac12\end{pmatrix}, \quad I = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}.$$

Hence
$$N = M - I = \begin{pmatrix}\dfrac52-1 & \dfrac32 \\[4pt] -\dfrac32 & -\dfrac12-1\end{pmatrix} = \begin{pmatrix}\dfrac32 & \dfrac32 \\[4pt] -\dfrac32 & -\dfrac32\end{pmatrix} = \dfrac32\begin{pmatrix}1 & 1 \\ -1 & -1\end{pmatrix}.$$

Check that $$N$$ is nilpotent of index 2:
Let $$A = \begin{pmatrix}1 & 1 \\ -1 & -1\end{pmatrix}$$. Then
$$A^2 = \begin{pmatrix}1 & 1 \\ -1 & -1\end{pmatrix} \begin{pmatrix}1 & 1 \\ -1 & -1\end{pmatrix} = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}.$$
Therefore $$N^2 = \left(\dfrac32\right)^2 A^2 = 0.$$

For any positive integer $$k$$, $$(I+N)^k = I + kN$$ because the binomial expansion stops at the linear term when $$N^2 = 0.$$

Thus
$$M^{2022} = (I+N)^{2022} = I + 2022N.$$

Compute $$2022N$$:
$$2022N = 2022 \times \begin{pmatrix}\dfrac32 & \dfrac32 \\[4pt] -\dfrac32 & -\dfrac32\end{pmatrix} = \begin{pmatrix}2022 \times \dfrac32 & 2022 \times \dfrac32 \\[4pt] -2022 \times \dfrac32 & -2022 \times \dfrac32\end{pmatrix} = \begin{pmatrix}3033 & 3033 \\[4pt] -3033 & -3033\end{pmatrix}.$$

Add $$I$$ to finish:
$$M^{2022} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} + \begin{pmatrix}3033 & 3033 \\ -3033 & -3033\end{pmatrix} = \begin{pmatrix}3034 & 3033 \\ -3033 & -3032\end{pmatrix}.$$

Hence $$M^{2022} = \begin{pmatrix} 3034 & 3033 \\ -3033 & -3032 \end{pmatrix}.$$

Option A which is: $$\begin{pmatrix} 3034 & 3033 \\ -3033 & -3032 \end{pmatrix}$$