Matrix `A` such that `A^2=2A-I ,w h e r eI` is the identity matrix, the for `ngeq2. A^n` is equal to `2^(n-1)A-(n-1)l` b. `2^(n-1)A-I` c. `n A-(n-1)l` d. `n A-I`

Options

A. `2^(n-1) A-(n-1)I`
B. `2^(n-1) A-I`
C. `nA-(n-1)I`
D. `nA-I`

Explanation

To solve the problem, we need to find the expression for \( A^n \) given that \( A^2 = 2A - I \), where \( I \) is the identity matrix, and \( n \geq 2 \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ A^2 = 2A - I \]

2. **Find \( A^3 \)**:
\[ A^3 = A \cdot A^2 \]
Substitute \( A^2 \):
\[ A^3 = A \cdot (2A - I) = 2A^2 - A \]
Now substitute \( A^2 \) again:
\[ A^3 = 2(2A - I) - A = 4A - 2I - A = 3A - 2I \]

3. **Find \( A^4 \)**:
\[ A^4 = A \cdot A^3 \]
Substitute \( A^3 \):
\[ A^4 = A \cdot (3A - 2I) = 3A^2 - 2A \]
Now substitute \( A^2 \) again:
\[ A^4 = 3(2A - I) - 2A = 6A - 3I - 2A = 4A - 3I \]

4. **Identify the pattern**:
From the calculations:
- \( A^2 = 2A - I \)
- \( A^3 = 3A - 2I \)
- \( A^4 = 4A - 3I \)

We can observe a pattern forming:
- For \( n = 2 \), \( A^2 = 2A - 1I \)
- For \( n = 3 \), \( A^3 = 3A - 2I \)
- For \( n = 4 \), \( A^4 = 4A - 3I \)

It appears that:
\[ A^n = nA - (n-1)I \]

5. **Generalize the expression**:
Thus, for \( n \geq 2 \):
\[ A^n = nA - (n-1)I \]

### Conclusion:
The correct expression for \( A^n \) is:
\[ A^n = nA - (n-1)I \]