If `A={:[(1,0),(1,1)]:},"then "A^(2008)` is equal to

Options

A. `{:[(2008,0),(1,1)]:}`
B. `{:[(1,0),(2008,1)]:}`
C. `{:[(1,0),(1,2008)]:}`
D. `2007{:[(1,0),(1,1)]:}`

Answer

To find \( A^{2008} \) where \( A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \), we can follow these steps:

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

We start by multiplying matrix \( A \) by itself:

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

Calculating the elements:

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

Thus, we have:

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

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

Next, we calculate \( A^3 \):

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

Calculating the elements:

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

Thus, we have:

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

### Step 3: Identify the Pattern

From the calculations, we can see a pattern emerging:

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

It appears that \( A^n = \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} \).

### Step 4: Generalize for \( A^{2008} \)

Using the pattern identified, we can generalize:

\[ A^{2008} = \begin{pmatrix} 1 & 0 \\ 2008 & 1 \end{pmatrix} \]

### Final Answer

Thus, the final result is:

\[ A^{2008} = \begin{pmatrix} 1 & 0 \\ 2008 & 1 \end{pmatrix} \]