Let A be a square matrix of order 3, `A^(T)` be the transpose matrix of matrix A and `"AA"^(T)=4I`. If `d=|(2A^(T)+"AA"^(T)+adjA)/(2)|, ` then the value of 12d is equal to `(|A|lt 0)`

Answer

To solve the given problem step by step, we will follow the mathematical principles and properties of matrices.

### Step 1: Understand the given condition
We are given that \( AA^T = 4I \), where \( A \) is a square matrix of order 3 and \( I \) is the identity matrix of order 3.

### Step 2: Find the determinant of \( A \)
From the equation \( AA^T = 4I \), we can take the determinant on both sides:
\[ \text{det}(AA^T) = \text{det}(4I) \]
Using the property of determinants, we have:
\[ \text{det}(AA^T) = \text{det}(A) \cdot \text{det}(A^T) = (\text{det}(A))^2 \]
The determinant of \( 4I \) is:
\[ \text{det}(4I) = 4^3 = 64 \]
Thus, we have:
\[ (\text{det}(A))^2 = 64 \]
Taking the square root gives:
\[ \text{det}(A) = \pm 8 \]
Since it is given that \( \text{det}(A) < 0 \), we conclude:
\[ \text{det}(A) = -8 \]

### Step 3: Find the adjoint of \( A \)
Using the relation between the adjoint and determinant, we have:
\[ \text{adj}(A) = \text{det}(A) \cdot A^{-1} \]
From the property of the inverse, we know:
\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \]
Thus:
\[ \text{adj}(A) = -8 A^{-1} \]

### Step 4: Calculate \( D \)
We need to find:
\[ D = \frac{2A^T + AA^T + \text{adj}(A)}{2} \]
Substituting \( AA^T = 4I \):
\[ D = \frac{2A^T + 4I + \text{adj}(A)}{2} \]
Now substituting \( \text{adj}(A) = -2A^T \) (derived from the previous steps):
\[ D = \frac{2A^T + 4I - 2A^T}{2} \]
This simplifies to:
\[ D = \frac{4I}{2} = 2I \]

### Step 5: Find the determinant of \( D \)
Now we calculate:
\[ |D| = |2I| = 2^3 = 8 \]

### Step 6: Calculate \( 12d \)
We need to find:
\[ 12d = 12 \times |D| = 12 \times 8 = 96 \]

### Final Result
Thus, the value of \( 12d \) is:
\[ \boxed{96} \]