If `A=[(1,0),(1//2,1)]` then `A^(50)` is

Options

A. `[(1,0),(0,50)]`
B. `[(1,0),(50,1)]`
C. `[(1,25),(0,1)]`
D. `[(1,0),(25,1)]`

Answer

To find \( A^{50} \) for the matrix \( A = \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \), we can follow these steps:

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

To find \( A^2 \), we multiply matrix \( A \) by itself:

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

Calculating the entries:

- 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, we have:

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

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

Next, we calculate \( A^3 \) by multiplying \( A^2 \) by \( A \):

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

Calculating the entries:

- 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, we have:

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

### Step 3: Generalize \( A^n \)

From the calculations, we can see a pattern emerging. The general form for \( A^n \) can be expressed as:

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

### Step 4: Calculate \( A^{50} \)

Now, substituting \( n = 50 \):

\[ A^{50} = \begin{pmatrix} 1 & 0 \\ \frac{50}{2} & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 25 & 1 \end{pmatrix} \]

### Final Result

Thus, the final result is:

\[ A^{50} = \begin{pmatrix} 1 & 0 \\ 25 & 1 \end{pmatrix} \]

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