`A=[(1,0,0),(0,alpha,beta),(0,beta,alpha)]` and`abs(2A)^3=2^21` then find `alpha (alpha, beta inI^+)`

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the value of \( \alpha \) given the matrix \( A \) and the determinant condition. Let's break it down step by step.

### Step 1: Write down the matrix and find its determinant
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{pmatrix} \]

To find the determinant of \( A \), we can use the formula for the determinant of a \( 3 \times 3 \) matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix, this simplifies to:
\[ \text{det}(A) = 1 \cdot (\alpha \cdot \alpha - \beta \cdot \beta) = \alpha^2 - \beta^2 \]

### Step 2: Use the given condition
We are given that:
\[ \text{abs}(2A)^3 = 2^{21} \]
This can be interpreted as:
\[ |2A|^3 = 2^{21} \]
Now, we know that:
\[ |2A| = 2^3 |A| = 8 |\text{det}(A)| \]
Thus:
\[ (8 |\text{det}(A)|)^3 = 2^{21} \]
This simplifies to:
\[ 512 |\text{det}(A)|^3 = 2^{21} \]
Since \( 512 = 2^9 \), we can write:
\[ 2^9 |\text{det}(A)|^3 = 2^{21} \]
Dividing both sides by \( 2^9 \):
\[ |\text{det}(A)|^3 = 2^{21 - 9} = 2^{12} \]
Taking the cube root:
\[ |\text{det}(A)| = 2^{12/3} = 2^4 = 16 \]

### Step 3: Set up the equation
From Step 1, we have:
\[ |\text{det}(A)| = |\alpha^2 - \beta^2| = 16 \]
This gives us two cases:
1. \( \alpha^2 - \beta^2 = 16 \)
2. \( \beta^2 - \alpha^2 = 16 \) (which we can ignore since \( \alpha, \beta \in \mathbb{I}^+ \))

### Step 4: Factor the equation
Using the difference of squares:
\[ \alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta) = 16 \]
Now we need to find pairs of integers \( (x, y) \) such that \( x \cdot y = 16 \) where \( x = \alpha + \beta \) and \( y = \alpha - \beta \).

### Step 5: Find integer pairs
The pairs of factors of 16 are:
- \( (1, 16) \)
- \( (2, 8) \)
- \( (4, 4) \)

### Step 6: Solve for \( \alpha \) and \( \beta \)
1. **For \( (1, 16) \)**:
\[ \alpha + \beta = 16, \quad \alpha - \beta = 1 \]
Adding these:
\[ 2\alpha = 17 \implies \alpha = 8.5 \quad (\text{not an integer}) \]

2. **For \( (2, 8) \)**:
\[ \alpha + \beta = 8, \quad \alpha - \beta = 2 \]
Adding these:
\[ 2\alpha = 10 \implies \alpha = 5 \]
Substituting back:
\[ 5 + \beta = 8 \implies \beta = 3 \]

3. **For \( (4, 4) \)**:
\[ \alpha + \beta = 4, \quad \alpha - \beta = 4 \]
Adding these:
\[ 2\alpha = 8 \implies \alpha = 4 \]
Substituting back:
\[ 4 + \beta = 4 \implies \beta = 0 \quad (\text{not positive}) \]

### Conclusion
The only valid solution is:
\[ \alpha = 5, \quad \beta = 3 \]

Thus, the value of \( \alpha \) is:
\[ \boxed{5} \]