If `A=[(1, 1),(1, 1)]`, then `A^(100)` is equal to

Options

A. `100A`
B. `2^(99)A`
C. `2^(100)A`
D. 99A

Answer

To find \( A^{100} \) where \( A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \), we can follow these steps:

### Step 1: Calculate \( A^2 \)
We start by calculating \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]
To perform the matrix multiplication, we calculate each element:
- First row, first column: \( 1 \cdot 1 + 1 \cdot 1 = 1 + 1 = 2 \)
- First row, second column: \( 1 \cdot 1 + 1 \cdot 1 = 1 + 1 = 2 \)
- Second row, first column: \( 1 \cdot 1 + 1 \cdot 1 = 1 + 1 = 2 \)
- Second row, second column: \( 1 \cdot 1 + 1 \cdot 1 = 1 + 1 = 2 \)

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

### Step 2: Calculate \( A^3 \)
Next, we calculate \( A^3 \):
\[ A^3 = A^2 \cdot A = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]
Calculating each element:
- First row, first column: \( 2 \cdot 1 + 2 \cdot 1 = 2 + 2 = 4 \)
- First row, second column: \( 2 \cdot 1 + 2 \cdot 1 = 2 + 2 = 4 \)
- Second row, first column: \( 2 \cdot 1 + 2 \cdot 1 = 2 + 2 = 4 \)
- Second row, second column: \( 2 \cdot 1 + 2 \cdot 1 = 2 + 2 = 4 \)

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

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

We can observe that:
\[ A^n = \begin{pmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{pmatrix} \]

### Step 4: Calculate \( A^{100} \)
Using the pattern identified, we can calculate \( A^{100} \):
\[ A^{100} = \begin{pmatrix} 2^{100-1} & 2^{100-1} \\ 2^{100-1} & 2^{100-1} \end{pmatrix} = \begin{pmatrix} 2^{99} & 2^{99} \\ 2^{99} & 2^{99} \end{pmatrix} \]

### Final Answer
Thus, the final answer is:
\[ A^{100} = \begin{pmatrix} 2^{99} & 2^{99} \\ 2^{99} & 2^{99} \end{pmatrix} \]