Let A be a matrix of order 3 such that `A^(2)=3A-2I` where, I is an identify matrix of order 3. If `A^(5)=alphaa+betaI`, then `alphabeta` is equal to

Options

A. 2025
B. `-2025`
C. `-930`
D. `-640`

Answer

To solve the problem, we need to find \( A^5 \) in terms of \( A \) and \( I \) using the given equation \( A^2 = 3A - 2I \).

### Step-by-Step Solution:

1. **Find \( A^3 \)**:
\[ A^3 = A^2 \cdot A \]
Substitute \( A^2 \) from the given equation:
\[ A^3 = (3A - 2I) \cdot A = 3A^2 - 2IA \]
Now substitute \( A^2 \) again:
\[ A^3 = 3(3A - 2I) - 2A = 9A - 6I - 2A = 7A - 6I \]

2. **Find \( A^4 \)**:
\[ A^4 = A^3 \cdot A \]
Substitute \( A^3 \):
\[ A^4 = (7A - 6I) \cdot A = 7A^2 - 6IA \]
Substitute \( A^2 \) again:
\[ A^4 = 7(3A - 2I) - 6A = 21A - 14I - 6A = 15A - 14I \]

3. **Find \( A^5 \)**:
\[ A^5 = A^4 \cdot A \]
Substitute \( A^4 \):
\[ A^5 = (15A - 14I) \cdot A = 15A^2 - 14IA \]
Substitute \( A^2 \) again:
\[ A^5 = 15(3A - 2I) - 14A = 45A - 30I - 14A = 31A - 30I \]

4. **Express \( A^5 \) in terms of \( \alpha A + \beta I \)**:
We have:
\[ A^5 = 31A - 30I \]
Thus, we can identify:
\[ \alpha = 31, \quad \beta = -30 \]

5. **Calculate \( \alpha \beta \)**:
\[ \alpha \beta = 31 \cdot (-30) = -930 \]

### Final Answer:
Thus, the value of \( \alpha \beta \) is \( -930 \).