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

Options

A. `2^100 A`
B. `2^99 A`
C. `2^101A`
D. None of these

Answer

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

### Step 1: Calculate \( A^2 \)
We start by calculating \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 1 \cdot 1 + 1 \cdot 1 = 2 \)
- First row, second column: \( 1 \cdot 1 + 1 \cdot 1 = 2 \)
- Second row, first column: \( 1 \cdot 1 + 1 \cdot 1 = 2 \)
- Second row, second column: \( 1 \cdot 1 + 1 \cdot 1 = 2 \)

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

### Step 2: Express \( A^2 \) in terms of \( A \)
We can factor out 2 from \( A^2 \):
\[ A^2 = 2 \cdot A \]

### Step 3: Calculate \( A^3 \)
Next, we find \( A^3 \):
\[ A^3 = A^2 \cdot A = (2A) \cdot A = 2A^2 \]
Substituting \( A^2 = 2A \):
\[ A^3 = 2 \cdot (2A) = 2^2 \cdot A \]

### Step 4: Generalize the pattern
Continuing this pattern, we can see:
- \( A^2 = 2^1 \cdot A \)
- \( A^3 = 2^2 \cdot A \)
- \( A^4 = 2^3 \cdot A \)

In general, we can express \( A^n \) as:
\[ A^n = 2^{n-1} \cdot A \]

### Step 5: Calculate \( A^{100} \)
Now, we apply this general formula for \( n = 100 \):
\[ A^{100} = 2^{100-1} \cdot A = 2^{99} \cdot A \]

### Final Result
Thus, the final result is:
\[ A^{100} = 2^{99} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]

### Conclusion
The answer is \( A^{100} = 2^{99} A \).