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 will analyze the matrices and their determinants as given in the question.

### Step 1: Define the matrices
Let \( A = [a_{ij}] \) be a square matrix of order 3. This means \( A \) is a \( 3 \times 3 \) matrix. We also have another matrix \( B = [b_{ij}] \) defined by the relation:
\[ b_{ij} = 2^{i-j} a_{ij} \quad \text{for } 1 \leq i,j \leq 3 \]

### Step 2: Write down the matrices
The matrix \( A \) can be represented as:
\[ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \]

Using the definition of \( B \), we can write:
\[ B = \begin{bmatrix} 2^{1-1} a_{11} & 2^{1-2} a_{12} & 2^{1-3} a_{13} \\ 2^{2-1} a_{21} & 2^{2-2} a_{22} & 2^{2-3} a_{23} \\ 2^{3-1} a_{31} & 2^{3-2} a_{32} & 2^{3-3} a_{33} \end{bmatrix} \]
This simplifies to:
\[ B = \begin{bmatrix} a_{11} & \frac{1}{2} a_{12} & \frac{1}{4} a_{13} \\ 2 a_{21} & a_{22} & \frac{1}{2} a_{23} \\ 4 a_{31} & 2 a_{32} & a_{33} \end{bmatrix} \]

### Step 3: Calculate the determinant of \( B \)
To find the determinant of \( B \), we can use the property of determinants with respect to row operations. If we multiply a row by a scalar, the determinant is multiplied by that scalar.

1. Multiply the first row by \( 4 \).
2. Multiply the second row by \( 2 \).
3. Multiply the third row by \( 1 \) (no change).

Thus, we have:
\[ \text{det}(B) = 4 \cdot 2 \cdot 1 \cdot \text{det}(A) = 8 \cdot \text{det}(A) \]

Given that \( \text{det}(A) = 3 \) (since the determinant of \( A \) is the same as its order), we find:
\[ \text{det}(B) = 8 \cdot 3 = 24 \]

### Step 4: Calculate the determinant of \( B^T \)
The determinant of the transpose of a matrix is equal to the determinant of the matrix itself:
\[ \text{det}(B^T) = \text{det}(B) = 24 \]

### Step 5: Calculate the determinant of \( (B^T)^{-1} \)
The determinant of the inverse of a matrix is the reciprocal of the determinant of the matrix:
\[ \text{det}((B^T)^{-1}) = \frac{1}{\text{det}(B^T)} = \frac{1}{24} \]

### Step 6: Find the magnitude of \( |(B^T)^{-1}| \)
Thus, the magnitude of \( |(B^T)^{-1}| \) is:
\[ |(B^T)^{-1}| = \frac{1}{24} \]

### Final Answer
The value of \( |(B^T)^{-1}| \) is \( \frac{1}{24} \).

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