Let for `A =[[1,2,3],[alpha,3,1],[1,1,2]], absA=2` . If `|2adj (2 adj (2A))|=32^n+alpha` is equal to

Options

A. `11`
B. `9`
C. `12`
D. `10`

Answer

To solve the problem step-by-step, we will first find the value of \( \alpha \) using the determinant of matrix \( A \), and then we will compute \( |2 \text{adj}(2 \text{adj}(2A))| \) to find \( n \) and finally calculate \( 3n + \alpha \).

### Step 1: Find the determinant of matrix \( A \)

Given:
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix} \]
We know that \( |A| = 2 \).

Using the determinant formula for a 3x3 matrix:
\[ |A| = 1 \cdot (3 \cdot 2 - 1 \cdot 1) - 2 \cdot (\alpha \cdot 2 - 1 \cdot 1) + 3 \cdot (\alpha \cdot 1 - 3 \cdot 1) \]
Calculating this:
\[ |A| = 1 \cdot (6 - 1) - 2 \cdot (2\alpha - 1) + 3 \cdot (\alpha - 3) \]
\[ = 5 - 4\alpha + 2 + 3\alpha - 9 \]
\[ = -4\alpha + 5 + 2 - 9 \]
\[ = -4\alpha - 2 \]
Setting this equal to 2:
\[ -4\alpha - 2 = 2 \]
\[ -4\alpha = 4 \]
\[ \alpha = -1 \]

### Step 2: Compute \( |2 \text{adj}(2 \text{adj}(2A))| \)

First, we need to compute \( 2A \):
\[ 2A = 2 \begin{bmatrix} 1 & 2 & 3 \\ -1 & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 6 \\ -2 & 6 & 2 \\ 2 & 2 & 4 \end{bmatrix} \]

Now, we find \( |2A| \):
\[ |2A| = 2^3 |A| = 8 \cdot 2 = 16 \]

Next, we calculate \( \text{adj}(2A) \):
\[ |\text{adj}(2A)| = |A|^{2} = 2^{2} = 4 \]

Now, we compute \( |2 \text{adj}(2A)| \):
\[ |2 \text{adj}(2A)| = 2^3 |\text{adj}(2A)| = 8 \cdot 4 = 32 \]

Next, we compute \( \text{adj}(2 \text{adj}(2A)) \):
\[ |\text{adj}(2 \text{adj}(2A))| = |2 \text{adj}(2A)|^{2} = 32^{2} = 1024 \]

Finally, we compute \( |2 \text{adj}(2 \text{adj}(2A))| \):
\[ |2 \text{adj}(2 \text{adj}(2A))| = 2^3 |\text{adj}(2 \text{adj}(2A))| = 8 \cdot 1024 = 8192 \]

### Step 3: Relate to the expression \( |2 \text{adj}(2 \text{adj}(2A))| = 32^n + \alpha \)

We know:
\[ 8192 = 32^n - 1 \]
Since \( 32 = 2^5 \), we can express \( 8192 \) as:
\[ 8192 = 2^{13} \]
Thus:
\[ 2^{13} = (2^5)^n - 1 \]
This implies:
\[ 2^{13} = 2^{5n} - 1 \]
So, \( 5n = 13 \) gives \( n = \frac{13}{5} \).

### Step 4: Calculate \( 3n + \alpha \)

Now substituting \( n \) and \( \alpha \):
\[ 3n + \alpha = 3 \cdot \frac{13}{5} - 1 = \frac{39}{5} - 1 = \frac{39}{5} - \frac{5}{5} = \frac{34}{5} \]

### Final Answer
Thus, the answer is:
\[ \boxed{11} \]