If $M=\left[\begin{array}{cc}\frac{1}{2}& \frac{3}{2}\\ \frac{3}{2}& \frac{5}{2}\end{array}\right]$

then which of the following matrices is equal to $M^{2022}$?

The correct answer is Option 1:

$M^{2022}=\left[\begin{array}{cc}3034& 3033\\ -3033& -3032\end{array}\right]$

We solve this using a key structural decomposition of the matrix $M$. The core idea is to write $M$ as the sum of the identity matrix and a nilpotent matrix, then exploit the resulting pattern to compute any power of $M$ with ease.

Step 1 — Decompose M into I + N

We observe that $M$ can be written as:

$M=I+\frac{3}{2}A$

where $I$ is the 2×2 identity matrix and $A=\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$.

Let $N=\frac{3}{2}A$, so that $M=I+N$.

Step 2 — Verify that N is nilpotent (N² = 0)

We compute $A^2$ directly:

$A^2=\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$

Computing each entry:
- Top-left: $1\times 1+1\times \left(-1\right)=0$
- Top-right: $1\times 1+1\times \left(-1\right)=0$
- Bottom-left: $\left(-1\right)\times 1+\left(-1\right)\times \left(-1\right)=0$
- Bottom-right: $\left(-1\right)\times 1+\left(-1\right)\times \left(-1\right)=0$

Therefore:

$A^2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

This means $N^2={\left(\frac{3}{2}A\right)}^2=\frac{9}{4}{A}^2=O$ (the zero matrix). So $N$ is nilpotent of order 2.

Step 3 — Apply the Binomial Theorem

Since $I$ and $N$ commute (as $I$ commutes with everything), we can expand:

$M^n={(I+N)}^n=\sum _{k=0}^n\left(\frac{n}{k}\right)N^k$

Because $N^2=O$, every term with $k\ge 2$ vanishes. Only the first two terms survive:

$M^n=N^0+n{N}^1=I+nN$

This gives us a beautiful, simple general formula:

$M^n=I+\frac{3n}{2}A=\left[\begin{array}{cc}1+\frac{3n}{2}& \frac{3n}{2}\\ -\frac{3n}{2}& 1-\frac{3n}{2}\end{array}\right]$

Step 4 — Substitute n = 2022

First compute the key coefficient:

$\frac{3\times 2022}{2}=3\times 1011=3033$

Now substitute into the general formula:

$M^{2022}=\left[\begin{array}{cc}1+3033& 3033\\ -3033& 1-3033\end{array}\right]=\left[\begin{array}{cc}3034& 3033\\ -3033& -3032\end{array}\right]$

This exactly matches Option 1. ✓

Quick Verification using determinant:
We can double-check using the determinant. The determinant of the answer matrix is:

$\left(3034\right)\left(-3032\right)-\left(3033\right)\left(-3033\right)=3033{}^2-(3033+1)(3033-1)=3033{}^2-(3033{}^2-1)=1$

Since $det\left(M\right)=1$, we have $det\left(M^{2022}\right)=1$. ✓ This confirms the result.