Matrix A is such that `A^(2)=2A-I`, where I is the identify matrix. Then for `n ne 2, A^(n)=`

Options

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

Answer

To solve the problem where matrix A satisfies the equation \( A^2 = 2A - I \), we want to find \( A^n \) for \( n \neq 2 \).

### 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 \( A^2 \) from the first equation:
\[ A^3 = A(2A - I) = 2A^2 - A \]
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 \( A^3 \):
\[ A^4 = A(3A - 2I) = 3A^2 - 2A \]
Substitute \( A^2 \):
\[ A^4 = 3(2A - I) - 2A = 6A - 3I - 2A = 4A - 3I \]

4. **Find \( A^5 \)**:
\[ A^5 = A \cdot A^4 \]
Substitute \( A^4 \):
\[ A^5 = A(4A - 3I) = 4A^2 - 3A \]
Substitute \( A^2 \):
\[ A^5 = 4(2A - I) - 3A = 8A - 4I - 3A = 5A - 4I \]

5. **Observe the pattern**:
From the calculations, we can see a pattern emerging:
- \( A^2 = 2A - I \)
- \( A^3 = 3A - 2I \)
- \( A^4 = 4A - 3I \)
- \( A^5 = 5A - 4I \)

6. **Generalize the formula**:
It appears that for \( n \geq 2 \):
\[ A^n = nA - (n-1)I \]

7. **Final Answer**:
Therefore, for \( n \neq 2 \):
\[ A^n = nA - (n-1)I \]