$$A=\left(\begin{array}{cc}\frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2}\end{array}\right)$$

$$A^{2}=\left(\begin{array}{cc}4 & 3 \\ -3 & -2\end{array}\right)$$

$$A^{3}=\left(\begin{array}{cc}\frac{11}{2} & \frac{9}{2} \\ -\frac{9}{2} & -\frac{7}{2}\end{array}\right)$$

$$A^{4}=\left(\begin{array}{cc}7 & 6 \\ -6 & -5\end{array}\right)$$

and so on

$A^n=\left(\begin{array}{cc}\frac{3 n}{2}+1 & \frac{3 n}{2} \\ -\frac{3 n}{2} & -\frac{3 n}{2}+1\end{array}\right)$

Now

$$ \begin{aligned} A^{2022} & =\left(\begin{array}{ll} \frac{3 \times 2022}{2}+1 & \frac{3 \times 2022}{2} \\ \frac{-3 \times 2022}{2} & \frac{-3 \times 2022}{2}+1 \end{array}\right) \\\\ & =\left(\begin{array}{cc} 3034 & 3033 \\ -3033 & -3032 \end{array}\right) \end{aligned} $$

$\therefore$ Option $(\mathrm{A})$ is correct