If `A = [{:(2,2),(2,2):}]`, then what is `A^(n)` equal to ?

Options

A. `[{:(2^(n),2^(n)),(2^(n),2^(n)):}]`
B. `[{:(2n,2n),(2n,2n):}]`
C. `[{:(2^(2n-1),2^(2n-1)),(2^(2n-1),2^(2n-1)):}]`
D. `[{:(2^(2n+1),2^(2n+1)),(2^(2n+1),2^(2n+1)):}]`

Explanation

To find \( A^n \) for the matrix \( A = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate \( A^2 \)
To find \( A^2 \), we multiply matrix \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \cdot \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 2 \cdot 2 + 2 \cdot 2 = 4 + 4 = 8 \)
- First row, second column: \( 2 \cdot 2 + 2 \cdot 2 = 4 + 4 = 8 \)
- Second row, first column: \( 2 \cdot 2 + 2 \cdot 2 = 4 + 4 = 8 \)
- Second row, second column: \( 2 \cdot 2 + 2 \cdot 2 = 4 + 4 = 8 \)

Thus,

\[ A^2 = \begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)
Now, we calculate \( A^3 \) by multiplying \( A^2 \) by \( A \):

\[ A^3 = A^2 \cdot A = \begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix} \cdot \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 8 \cdot 2 + 8 \cdot 2 = 16 + 16 = 32 \)
- First row, second column: \( 8 \cdot 2 + 8 \cdot 2 = 16 + 16 = 32 \)
- Second row, first column: \( 8 \cdot 2 + 8 \cdot 2 = 16 + 16 = 32 \)
- Second row, second column: \( 8 \cdot 2 + 8 \cdot 2 = 16 + 16 = 32 \)

Thus,

\[ A^3 = \begin{pmatrix} 32 & 32 \\ 32 & 32 \end{pmatrix} \]

### Step 3: Identify the Pattern
From the calculations, we can observe a pattern:

- \( A^1 = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} = 2 \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \)
- \( A^2 = \begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix} = 2^3 \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \)
- \( A^3 = \begin{pmatrix} 32 & 32 \\ 32 & 32 \end{pmatrix} = 2^5 \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \)

### Step 4: Generalize the Formula
From the observed pattern, we can generalize:

\[ A^n = 2^{2n-1} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]

### Final Answer
Thus, the final result is:

\[ A^n = 2^{2n-1} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]