Let `A = [{:(1,1,1),(1,1,1),(1,1,1):}]` be a square matrix of order 3. Then for any positive integer n, what is `A^(n)` equal to ?

Options

A. A
B. `3^(n)A`
C. `(3^(n-1))A`
D. 3A

Explanation

To solve the problem, we need to find \( A^n \) for the matrix \( A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \) for any positive integer \( n \).

### Step-by-Step Solution:

1. **Identify the Matrix**:
The given matrix \( A \) is:
\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \]

2. **Calculate \( A^2 \)**:
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \]
Each element of the resulting matrix is the sum of the products of the corresponding row and column:
- For the first row and first column: \( 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = 3 \)
- This is true for all elements in the resulting matrix.

Thus,
\[ A^2 = \begin{pmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \end{pmatrix} = 3A \]

3. **Calculate \( A^3 \)**:
Now we calculate \( A^3 \) using \( A^2 \):
\[ A^3 = A^2 \cdot A = (3A) \cdot A = 3A^2 \]
From the previous step, we know \( A^2 = 3A \):
\[ A^3 = 3(3A) = 9A \]

4. **Generalize for \( A^n \)**:
From the calculations above, we observe a pattern:
- \( A^1 = A \)
- \( A^2 = 3A \)
- \( A^3 = 9A = 3^2 A \)

We can generalize this pattern:
\[ A^n = 3^{n-1} A \]

### Final Result:
Thus, for any positive integer \( n \):
\[ A^n = 3^{n-1} A \]