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 start with the equation given for the matrix \( A \):

\[ A^2 = 3A - 2I \]

where \( I \) is the identity matrix of order 3. We need to find \( A^5 \) in the form \( A^5 = \alpha A + \beta I \) and then compute the product \( \alpha \beta \).

### Step 1: Find \( A^3 \)

We can express \( A^3 \) using \( A^2 \):

\[ A^3 = A \cdot A^2 \]

Substituting \( A^2 \):

\[ A^3 = A(3A - 2I) = 3A^2 - 2A \]

Now, we substitute \( A^2 \) again:

\[ A^3 = 3(3A - 2I) - 2A = 9A - 6I - 2A = 7A - 6I \]

### Step 2: Find \( A^4 \)

Next, we find \( A^4 \):

\[ A^4 = A \cdot A^3 \]

Substituting \( A^3 \):

\[ A^4 = A(7A - 6I) = 7A^2 - 6A \]

Substituting \( A^2 \):

\[ A^4 = 7(3A - 2I) - 6A = 21A - 14I - 6A = 15A - 14I \]

### Step 3: Find \( A^5 \)

Now, we find \( A^5 \):

\[ A^5 = A \cdot A^4 \]

Substituting \( A^4 \):

\[ A^5 = A(15A - 14I) = 15A^2 - 14A \]

Substituting \( A^2 \):

\[ A^5 = 15(3A - 2I) - 14A = 45A - 30I - 14A = 31A - 30I \]

### Step 4: Identify \( \alpha \) and \( \beta \)

From the expression we derived for \( A^5 \):

\[ A^5 = 31A - 30I \]

We can identify \( \alpha = 31 \) and \( \beta = -30 \).

### Step 5: Calculate \( \alpha \beta \)

Now we calculate the product \( \alpha \beta \):

\[ \alpha \beta = 31 \times (-30) = -930 \]

Thus, the final answer is:

\[ \alpha \beta = -930 \]