If A and B are two non-singular matrices of order 3 such that `A A^(T)=2I` and `A^(-1)=A^(T)-A`. Adj. `(2B^(-1))`, then det. (B) is equal to

Options

A. `4`
B. `4sqrt(2)`
C. `16`
D. `16sqrt(2)`

Explanation

To solve the given problem step by step, we will analyze the conditions provided and apply properties of matrices.

### Step 1: Analyze the given conditions
We have two non-singular matrices \( A \) and \( B \) of order 3 such that:
1. \( A A^T = 2I \)
2. \( A^{-1} = A^T - A \cdot \text{Adj}(2B^{-1}) \)

### Step 2: Use the first condition
From the first condition \( A A^T = 2I \), we can take the determinant of both sides:
\[ \text{det}(A A^T) = \text{det}(2I) \]
Using the property of determinants, we have:
\[ \text{det}(A) \cdot \text{det}(A^T) = \text{det}(2I) \]
Since \( \text{det}(A^T) = \text{det}(A) \), we can write:
\[ \text{det}(A)^2 = 2^3 = 8 \]
Thus,
\[ \text{det}(A) = \sqrt{8} = 2\sqrt{2} \]

### Step 3: Use the second condition
Now, we analyze the second condition \( A^{-1} = A^T - A \cdot \text{Adj}(2B^{-1}) \). Multiply both sides by \( A \):
\[ I = A A^T - A^2 \cdot \text{Adj}(2B^{-1}) \]
Substituting \( A A^T = 2I \):
\[ I = 2I - A^2 \cdot \text{Adj}(2B^{-1}) \]
Rearranging gives:
\[ A^2 \cdot \text{Adj}(2B^{-1}) = I \]

### Step 4: Take determinants
Taking the determinant of both sides:
\[ \text{det}(A^2) \cdot \text{det}(\text{Adj}(2B^{-1})) = \text{det}(I) \]
Since \( \text{det}(I) = 1 \) and \( \text{det}(A^2) = (\text{det}(A))^2 = 8 \):
\[ 8 \cdot \text{det}(\text{Adj}(2B^{-1})) = 1 \]
Using the property of adjugates, we have:
\[ \text{det}(\text{Adj}(2B^{-1})) = (\text{det}(2B^{-1}))^2 \]
Thus,
\[ 8 \cdot (\text{det}(2B^{-1}))^2 = 1 \]

### Step 5: Find \( \text{det}(2B^{-1}) \)
We know that:
\[ \text{det}(2B^{-1}) = 2^3 \cdot \text{det}(B^{-1}) = 8 \cdot \frac{1}{\text{det}(B)} \]
Substituting this into our equation:
\[ 8 \cdot \left(8 \cdot \frac{1}{\text{det}(B)}\right)^2 = 1 \]
This simplifies to:
\[ 8 \cdot 64 \cdot \frac{1}{(\text{det}(B))^2} = 1 \]
\[ \frac{512}{(\text{det}(B))^2} = 1 \]
Thus,
\[ (\text{det}(B))^2 = 512 \]
Taking the square root gives:
\[ \text{det}(B) = \sqrt{512} = 16\sqrt{2} \]

### Final Answer
Therefore, the determinant of \( B \) is:
\[ \text{det}(B) = 16\sqrt{2} \]