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 problem step by step, we will follow the given conditions and manipulate the equations accordingly.

### Step 1: Understand the given conditions
We are given 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: Find the determinant of \( A \)
From the first condition \( A A^T = 2I \), we can take the determinant on both sides.

\[ \det(A A^T) = \det(2I) \]

Using the property of determinants, we have:

\[ \det(A) \cdot \det(A^T) = \det(2I) \]

Since \( \det(A^T) = \det(A) \), we can write:

\[ \det(A)^2 = \det(2I) \]

The determinant of \( 2I \) for a 3x3 matrix is:

\[ \det(2I) = 2^3 = 8 \]

Thus, we have:

\[ \det(A)^2 = 8 \]

Taking the square root, we find:

\[ \det(A) = \pm 2\sqrt{2} \]

### Step 3: Use the second condition
Now, we use the second condition \( A^{-1} = A^T - A \cdot \text{Adj}(2B^{-1}) \). Pre-multiplying both sides by \( A \):

\[ I = A A^T - A^2 \cdot \text{Adj}(2B^{-1}) \]

From the first condition, we know \( A A^T = 2I \):

\[ I = 2I - A^2 \cdot \text{Adj}(2B^{-1}) \]

Rearranging gives:

\[ A^2 \cdot \text{Adj}(2B^{-1}) = I \]

### Step 4: Take the determinant
Taking the determinant of both sides:

\[ \det(A^2 \cdot \text{Adj}(2B^{-1})) = \det(I) \]

Since \( \det(I) = 1 \), we have:

\[ \det(A^2) \cdot \det(\text{Adj}(2B^{-1})) = 1 \]

We already found \( \det(A^2) = 8 \), so:

\[ 8 \cdot \det(\text{Adj}(2B^{-1})) = 1 \]

Thus:

\[ \det(\text{Adj}(2B^{-1})) = \frac{1}{8} \]

### Step 5: Use the property of adjoint
Using the property of adjoint, we know:

\[ \det(\text{Adj}(X)) = \det(X)^{n-1} \quad \text{for an } n \times n \text{ matrix} \]

For \( n = 3 \):

\[ \det(\text{Adj}(2B^{-1})) = \det(2B^{-1})^2 \]

Calculating \( \det(2B^{-1}) \):

\[ \det(2B^{-1}) = 2^3 \cdot \det(B^{-1}) = 8 \cdot \frac{1}{\det(B)} \]

Thus:

\[ \det(\text{Adj}(2B^{-1})) = \left(8 \cdot \frac{1}{\det(B)}\right)^2 = \frac{64}{\det(B)^2} \]

### Step 6: Set the equations equal
We have:

\[ \frac{64}{\det(B)^2} = \frac{1}{8} \]

Cross-multiplying gives:

\[ 64 = \frac{1}{8} \cdot \det(B)^2 \]

Multiplying both sides by 8:

\[ 512 = \det(B)^2 \]

Taking the square root gives:

\[ \det(B) = \pm 16\sqrt{2} \]

### Final Answer
Thus, the value of \( \det(B) \) is \( 16\sqrt{2} \) or \( -16\sqrt{2} \).

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