If $M = \begin{pmatrix} \dfrac{5}{2} & \dfrac{3}{2} \\[6pt] -\dfrac{3}{2} & -\dfrac{1}{2} \end{pmatrix}$, then which of the following matrices is equal to $M^{2022}$?

$M = \begin{pmatrix} \tfrac{5}{2} & \tfrac{3}{2} \\ -\tfrac{3}{2} & -\tfrac{1}{2} \end{pmatrix}$ $\text{To find: } M^{2022}$ $1.\ \operatorname{tr}(M), \det(M) \Rightarrow \text{characteristic equation}$ $2.\ \text{show } (M-I)^2 = O$ $3.\ M^n = (I + N)^n = I + nN, \quad N = M - I$ $\operatorname{tr}(M) = \dfrac{5}{2}-\dfrac{1}{2} = 2, \qquad \det(M) = \dfrac{5}{2}\left(-\dfrac{1}{2}\right) - \dfrac{3}{2}\left(-\dfrac{3}{2}\right) = 1$ $\lambda^2 - 2\lambda + 1 = 0 \Rightarrow (\lambda-1)^2 = 0$ $(M-I)^2 = O \ \Rightarrow \ M = I + N, \quad N = M - I, \quad N^2 = O$ $N = M - I = \begin{pmatrix} \tfrac{3}{2} & \tfrac{3}{2} \\ -\tfrac{3}{2} & -\tfrac{3}{2} \end{pmatrix} = \dfrac{3}{2}\begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$ $N^2 = \dfrac{9}{4}\begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}^2 = O$ $M^n = (I+N)^n = I + nN \quad (\text{since } N^2 = O)$ $M^n = I + nN = \begin{pmatrix} 1 + \tfrac{3n}{2} & \tfrac{3n}{2} \\[2pt] -\tfrac{3n}{2} & 1 - \tfrac{3n}{2} \end{pmatrix}$ $n = 2022: \quad \dfrac{3n}{2} = \dfrac{3 \times 2022}{2} = 3033$ $M^{2022} = \begin{pmatrix} 3034 & 3033 \\ -3033 & -3032 \end{pmatrix} \ \Rightarrow \ \text{option } (1)$