Let `A=[a_(ij)]` be a square matrix of order 3 and `B=[b_(ij)]` be a matrix such that `b_(ij)=2^(i-j)a_(ij)` for ` 1lei,jle3, AA i,j in N`.
If the determinant of A is same as its order, then the value of `|(B^(T))^(-1)|` is

Options

A. `(1)/(3)`
B. 3
C. 9
D. `(1)/(27)`

Answer

To solve the problem step by step, we need to find the value of \(|(B^T)^{-1}|\) given that the determinant of matrix \(A\) is equal to its order, which is 3.

### Step 1: Understand the matrices \(A\) and \(B\)

Given:
- \(A = [a_{ij}]\) is a square matrix of order 3.
- \(B = [b_{ij}]\) where \(b_{ij} = 2^{i-j} a_{ij}\).

### Step 2: Write out the matrix \(B\)

For \(i, j = 1, 2, 3\):
- \(b_{11} = 2^{1-1} a_{11} = a_{11}\)
- \(b_{12} = 2^{1-2} a_{12} = \frac{1}{2} a_{12}\)
- \(b_{13} = 2^{1-3} a_{13} = \frac{1}{4} a_{13}\)

- \(b_{21} = 2^{2-1} a_{21} = 2 a_{21}\)
- \(b_{22} = 2^{2-2} a_{22} = a_{22}\)
- \(b_{23} = 2^{2-3} a_{23} = \frac{1}{2} a_{23}\)

- \(b_{31} = 2^{3-1} a_{31} = 4 a_{31}\)
- \(b_{32} = 2^{3-2} a_{32} = 2 a_{32}\)
- \(b_{33} = 2^{3-3} a_{33} = a_{33}\)

Thus, the matrix \(B\) looks like:
\[ B = \begin{bmatrix} a_{11} & \frac{1}{2} a_{12} & \frac{1}{4} a_{13} \\ 2 a_{21} & a_{22} & \frac{1}{2} a_{23} \\ 4 a_{31} & 2 a_{32} & a_{33} \end{bmatrix} \]

### Step 3: Calculate the determinant of \(B\)

Using the property of determinants, we can factor out the constants from each row:
\[ |B| = 1 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 2 \cdot 1 \cdot \frac{1}{2} \cdot 4 \cdot 2 \cdot 1 \cdot |A| \]
Calculating the constants:
\[ |B| = \left(1 \cdot 1 \cdot 1\right) \cdot \left(\frac{1}{2} \cdot 2\right) \cdot \left(\frac{1}{4} \cdot 4\right) \cdot |A| = 1 \cdot 1 \cdot |A| = |A| \]

### Step 4: Use the given condition

Since it is given that \(|A| = 3\), we have:
\[ |B| = |A| = 3 \]

### Step 5: Find the determinant of \(B^T\)

The determinant of the transpose of a matrix is equal to the determinant of the matrix itself:
\[ |B^T| = |B| = 3 \]

### Step 6: Calculate the determinant of \((B^T)^{-1}\)

The determinant of the inverse of a matrix is the reciprocal of the determinant of the matrix:
\[ |(B^T)^{-1}| = \frac{1}{|B^T|} = \frac{1}{3} \]

### Final Answer

Thus, the value of \(|(B^T)^{-1}|\) is:
\[ \frac{1}{3} \]