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:

$\left[\begin{array}{cc}3032& 3033\\ 3033& 3034\end{array}\right]$

Step-by-Step Explanation:

We are given the matrix:

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

To find $M^{2022}$, let's analyze the linear pattern for powers of $M$.

Notice that each element of $M^n$ increases linearly with respect to $n$ by adding $\frac{3}{2}$ per step to every entry of the matrix for each incremental power.

The general formula for $M^n$ is given by:

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

For $n=2022$, we substitute $n=2022$ into the equation:

$(2022-1)·\frac{3}{2}=2021·\frac{3}{2}=\frac{6063}{2}=3031.5$

Now, add $3031.5$ (or $\frac{6063}{2}$) to each corresponding element of matrix $M$:

1. Row 1, Column 1:
$\frac{1}{2}+3031.5=0.5+3031.5=3032$

2. Row 1, Column 2:
$\frac{3}{2}+3031.5=1.5+3031.5=3033$

3. Row 2, Column 1:
$\frac{3}{2}+3031.5=1.5+3031.5=3033$

4. Row 2, Column 2:
$\frac{5}{2}+3031.5=2.5+3031.5=3034$

Combining these calculated elements into the resulting matrix gives:

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