If `A=[(1,0),(1//2,1)]`, `A^(400)` is equal to

Options

A. `((1,0),(50,1))`
B. `((1,0),((1//2)^(100),1))`
C. `((1,0),(25,1))`
D. None of these

Answer

To find \( A^{400} \) for the matrix \( A = \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \), we will first calculate \( A^2 \), then \( A^3 \), and look for a pattern to generalize \( A^n \).

### Step 1: Calculate \( A^2 \)

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 1 \cdot 1 + 0 \cdot \frac{1}{2} = 1 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 1 = 0 \)
- Second row, first column: \( \frac{1}{2} \cdot 1 + 1 \cdot \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1 \)
- Second row, second column: \( \frac{1}{2} \cdot 0 + 1 \cdot 1 = 0 + 1 = 1 \)

Thus,

\[ A^2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)

\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 1 \cdot 1 + 0 \cdot \frac{1}{2} = 1 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 1 = 0 \)
- Second row, first column: \( 1 \cdot 1 + 1 \cdot \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2} \)
- Second row, second column: \( 1 \cdot 0 + 1 \cdot 1 = 0 + 1 = 1 \)

Thus,

\[ A^3 = \begin{pmatrix} 1 & 0 \\ \frac{3}{2} & 1 \end{pmatrix} \]

### Step 3: Calculate \( A^4 \)

\[ A^4 = A^3 \cdot A = \begin{pmatrix} 1 & 0 \\ \frac{3}{2} & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 1 \cdot 1 + 0 \cdot \frac{1}{2} = 1 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 1 = 0 \)
- Second row, first column: \( \frac{3}{2} \cdot 1 + 1 \cdot \frac{1}{2} = \frac{3}{2} + \frac{1}{2} = 2 \)
- Second row, second column: \( \frac{3}{2} \cdot 0 + 1 \cdot 1 = 0 + 1 = 1 \)

Thus,

\[ A^4 = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \]

### Step 4: Identify the Pattern

From our calculations, we can see a pattern forming:

- \( A^1 = \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \)
- \( A^2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \)
- \( A^3 = \begin{pmatrix} 1 & 0 \\ \frac{3}{2} & 1 \end{pmatrix} \)
- \( A^4 = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \)

The pattern for the lower left element seems to be increasing by \( \frac{1}{2} \) for each power of \( A \):

- \( A^n = \begin{pmatrix} 1 & 0 \\ \frac{n}{2} & 1 \end{pmatrix} \)

### Step 5: Generalize for \( A^{400} \)

Using the identified pattern:

\[ A^{400} = \begin{pmatrix} 1 & 0 \\ \frac{400}{2} & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 200 & 1 \end{pmatrix} \]

### Final Answer

\[ A^{400} = \begin{pmatrix} 1 & 0 \\ 200 & 1 \end{pmatrix} \]