If `A=[{:(1,a),(0,1):}]` then `A^(n)` (where `n in N`) is equal to

Options

A. `[{:(1,na),(0,1):}]`
B. `[{:(1,na),(1,0):}]`
C. `[{:(n,na),(0,n):}]`
D. `[{:(1,n^(2)a),(0,1):}]`

Answer

To find \( A^n \) for the matrix \( A = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} \), we will calculate the powers of \( A \) step by step.

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

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

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

Calculating the elements:

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

Thus,

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

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

Now we calculate \( A^3 = A^2 \cdot A \):

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

Calculating the elements:

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

Thus,

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

### Step 3: Identify the Pattern

From the calculations, we can see a pattern emerging:

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

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

### Step 4: Generalize the Result

Thus, we can generalize that:

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

### Final Answer

The final result is:

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