If M = begin{bmatrix} frac{5}{2} & frac{3}{2} | vedclass

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(A) Given $M = \begin{bmatrix} \frac{5}{2} & \frac{3}{2} \ -\frac{3}{2} & -\frac{1}{2} \end{bmatrix}$.We can write $M$ as $M = I + \frac{3}{2} A$,whe

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$\begin{bmatrix} 3034 & 3033 \ -3033 & -3032 \end{bmatrix}$

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(A) Given $M = \begin{bmatrix} \frac{5}{2} & \frac{3}{2} \ -\frac{3}{2} & -\frac{1}{2} \end{bmatrix}$.We can write $M$ as $M = I + \frac{3}{2} A$,where $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ and $A = \begin{bmatrix} 1 & 1 \ -1 & -1 \end{bmatrix}$.First,calculate $A^2$:$A^2 = \begin{bmatrix} 1 & 1 \ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ -1 & -1 \end{bmatrix} = \begin{bmatrix} 1-1 & 1-1 \ -1+1 & -1+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} = O$.Since $A^2 = O$,we can use the binomial expansion for $(I + \frac{3}{2} A)^n$:$M^n = (I + \frac{3}{2} A)^n = I^n + n(I^{n-1})(\frac{3}{2} A) + \frac{n(n-1)}{2} I^{n-2} (\frac{3}{2} A)^2 + \dots$Since $A^2 = O$,all terms involving $A^k$ for $k \ge 2$ are zero.Thus,$M^n = I + n \cdot \frac{3}{2} A$.For $n = 2022$:$M^{2022} = I + 2022 \cdot \frac{3}{2} A = I + 3033 A$.$M^{2022} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + 3033 \begin{bmatrix} 1 & 1 \ -1 & -1 \end{bmatrix} = \begin{bmatrix} 1+3033 & 0+3033 \ 0-3033 & 1-3033 \end{bmatrix} = \begin{bmatrix} 3034 & 3033 \ -3033 & -3032 \end{bmatrix}$.

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

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