If `A=[{:(1,1),(1,1):}] and n in N` then `A^(n)` is qual to

Options

A. nA
B. 2nA
C. `2^(n-1)A`
D. `2^(n)A`

Answer

To solve the problem, we need to find \( A^n \) where \( A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \) and \( n \) is a natural number.

### Step-by-Step Solution:

1. **Calculate \( A^2 \)**:
\[ A^2 = A \times A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]
To compute this, we multiply the rows of the first matrix by the columns of the second matrix:
- First row, first column: \( 1 \times 1 + 1 \times 1 = 2 \)
- First row, second column: \( 1 \times 1 + 1 \times 1 = 2 \)
- Second row, first column: \( 1 \times 1 + 1 \times 1 = 2 \)
- Second row, second column: \( 1 \times 1 + 1 \times 1 = 2 \)

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

2. **Calculate \( A^3 \)**:
\[ A^3 = A^2 \times A = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \times \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]
Again, we multiply:
- First row, first column: \( 2 \times 1 + 2 \times 1 = 4 \)
- First row, second column: \( 2 \times 1 + 2 \times 1 = 4 \)
- Second row, first column: \( 2 \times 1 + 2 \times 1 = 4 \)
- Second row, second column: \( 2 \times 1 + 2 \times 1 = 4 \)

Thus,
\[ A^3 = \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix} \]

3. **Identify the pattern**:
From the calculations, we can see a pattern emerging:
- \( A^1 = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \)
- \( A^2 = 2 \cdot A \)
- \( A^3 = 4 \cdot A \)

This suggests that:
\[ A^n = 2^{n-1} \cdot A \]

4. **Generalize the formula**:
Therefore, we can express \( A^n \) as:
\[ A^n = 2^{n-1} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]

5. **Final result**:
Thus, the final answer is:
\[ A^n = 2^{n-1} \cdot A \]