If $$A = \begin{bmatrix}1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix}$$. Then $$A^{2022}$$ is

Let X be a matrix : $$\begin{bmatrix}a_1&a_2\\b_1&b_2\\\end{bmatrix}$$

Then $$X^2 = \begin{bmatrix}a_1^2+a_2b_1&a_1a_2+a_2b_2\\a_1b_1+b_1b_2&b_2^2+a_2b_1\\\end{bmatrix}$$

For the given matrix $$A = \begin{bmatrix}1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix}$$.

$$A^2 = \begin{bmatrix}1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix}$$.

This implies that $$A^2 = A$$, $$A^3 = A$$ , $$A^4 = A$$......and so on,....

Therefore $$A^{2022}=A$$