High Power Evaluation of Two by Two Matrix

To find the matrix $M^{2022}$, we first analyze the properties of the given matrix:

$M = \begin{pmatrix} \frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2} \end{pmatrix}$

Step 1: Compute the trace and determinant of $M$

• $\text{tr}(M) = \frac{5}{2} + \left(-\frac{1}{2}\right) = 2$
• $\det(M) = \left(\frac{5}{2}\right)\left(-\frac{1}{2}\right) - \left(\frac{3}{2}\right)\left(-\frac{3}{2}\right) = -\frac{5}{4} + \frac{9}{4} = 1$

Step 2: Find the characteristic equation of $M$ The characteristic polynomial of $M$ is given by: $\lambda^2 - \text{tr}(M)\lambda + \det(M) = 0$ $\lambda^2 - 2\lambda + 1 = 0 \implies (\lambda - 1)^2 = 0$

By the Cayley-Hamilton theorem, every matrix satisfies its own characteristic equation: $(M - I)^2 = 0$

Step 3: Decomposition into an identity matrix and a nilpotent matrix Let $N = M - I$. Then: $N = \begin{pmatrix} \frac{5}{2} - 1 & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2} - 1 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{3}{2} \end{pmatrix}$

Since $(M - I)^2 = 0$, $N$ is a nilpotent matrix of index 2, meaning $N^2 = 0$.

Step 4: Compute $M^{2022}$ using the binomial theorem Since $M = I + N$ and the identity matrix $I$ commutes with $N$: $M^n = (I + N)^n = \sum_{k=0}^n \binom{n}{k} I^{n-k} N^k$

Since $N^k = 0$ for all $k \ge 2$, the expansion simplifies to: $M^n = I + nN$

Substituting $n = 2022$: $M^{2022} = I + 2022 N$

Step 5: Perform the matrix addition $2022 N = 2022 \begin{pmatrix} \frac{3}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{3}{2} \end{pmatrix} = \begin{pmatrix} 3033 & 3033 \\ -3033 & -3033 \end{pmatrix}$

Now, adding the identity matrix $I$: $M^{2022} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 3033 & 3033 \\ -3033 & -3033 \end{pmatrix} = \begin{pmatrix} 3034 & 3033 \\ -3033 & -3032 \end{pmatrix}$

Thus, the correct option is (A).