If `A = [{:(1,1),(1,1):}] and A ^(100) = 2k^(k) A,` then the vlaue of k is

Answer

To solve the problem, we need to find the value of \( k \) given that \( A^{100} = 2^{k} A \).

### Step-by-Step Solution:

1. **Understanding the Matrix \( A \)**:
We are given \( A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \). This is a 2x2 matrix.

2. **Finding \( A^2 \)**:
We first calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \]
We can express this as:
\[ A^2 = 2A \]

3. **Finding a Pattern**:
Let's assume that \( A^n = 2^{n-1} A \) for some integer \( n \). We will prove this by induction.

4. **Base Case**:
For \( n = 1 \):
\[ A^1 = A = 2^{1-1} A = 2^0 A \]
The base case holds.

5. **Inductive Step**:
Assume it holds for \( n = k \):
\[ A^k = 2^{k-1} A \]
Now we need to show it holds for \( n = k + 1 \):
\[ A^{k+1} = A^k \cdot A = (2^{k-1} A) \cdot A = 2^{k-1} A^2 \]
From our earlier calculation, \( A^2 = 2A \), so:
\[ A^{k+1} = 2^{k-1} \cdot 2A = 2^k A \]
Thus, \( A^{k+1} = 2^k A \) holds.

6. **Conclusion of Induction**:
By induction, we have shown that \( A^n = 2^{n-1} A \) for all \( n \geq 1 \).

7. **Finding \( A^{100} \)**:
Now, substituting \( n = 100 \):
\[ A^{100} = 2^{100-1} A = 2^{99} A \]

8. **Setting Up the Equation**:
We know from the problem statement that:
\[ A^{100} = 2^k A \]
Therefore, we can equate:
\[ 2^{99} A = 2^k A \]

9. **Solving for \( k \)**:
Since \( A \neq 0 \), we can divide both sides by \( A \):
\[ 2^{99} = 2^k \]
This implies:
\[ k = 99 \]

### Final Answer:
The value of \( k \) is \( 99 \).