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

Options

A. `2^(100)A`
B. `2^(99)A`
C. `100A`
D. `299A`

Explanation

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

### Step 1: Calculate \( A^2 \)
To find \( A^2 \), we multiply matrix \( A \) by itself:

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

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,

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

### Step 2: Calculate \( A^4 \)
Now, we can find \( A^4 \) by squaring \( A^2 \):

\[ A^4 = A^2 \cdot A^2 = (2A) \cdot (2A) = 4A^2 \]

Substituting \( A^2 = 2A \):

\[ A^4 = 4(2A) = 8A \]

### Step 3: Generalize \( A^{2^n} \)
From the calculations, we can see a pattern:

- \( A^2 = 2A \)
- \( A^4 = 8A = 2^3 A \)

We can generalize this for \( A^{2^n} \):

\[ A^{2^n} = 2^{n+1} A \]

### Step 4: Calculate \( A^{100} \)
We can express \( 100 \) as \( 2^6 + 36 \) (since \( 100 = 64 + 36 \)), which means we can write:

\[ A^{100} = A^{64} \cdot A^{36} \]

Using our general formula for \( A^{64} \):

\[ A^{64} = 2^{64/2 + 1} A = 2^{33} A \]

Now we need to find \( A^{36} \). We can express \( 36 \) as \( 2^5 + 4 \) (since \( 36 = 32 + 4 \)):

\[ A^{36} = A^{32} \cdot A^4 \]

Using our general formula for \( A^{32} \):

\[ A^{32} = 2^{32/2 + 1} A = 2^{17} A \]

Thus,

\[ A^{36} = (2^{17} A) \cdot (8A) = 2^{17} \cdot 8 A^2 = 2^{17} \cdot 2^3 A = 2^{20} A \]

### Final Calculation
Now substituting back:

\[ A^{100} = A^{64} \cdot A^{36} = (2^{33} A) \cdot (2^{20} A) = 2^{33 + 20} A^2 = 2^{53} A^2 \]

Since \( A^2 = 2A \):

\[ A^{100} = 2^{53} \cdot 2A = 2^{54} A \]

So, the final result is:

\[ A^{100} = 2^{54} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]

### Summary
Thus, \( A^{100} = 2^{54} A \).