Let `A = [a_(ij)], B=[b_(ij)]` are two 3 × 3 matrices such that `b_(ij) = lambda ^(i+j-2) a_(ij)` & `|B| = 81`. Find |A| if `lambda` = 3.

Options

A. `1//81`
B. `1//9`
C. 3
D. `1//3`

Answer

To solve the problem, we need to find the determinant of matrix \( A \) given that the determinant of matrix \( B \) is 81 and the relationship between the elements of the two matrices is defined as:

\[ b_{ij} = \lambda^{i+j-2} a_{ij} \]

where \( \lambda = 3 \).

### Step-by-Step Solution:

1. **Understanding the relationship between matrices \( A \) and \( B \)**:
The elements of matrix \( B \) can be expressed in terms of the elements of matrix \( A \) as follows:
\[ b_{ij} = 3^{i+j-2} a_{ij} \]

2. **Formulating the matrices**:
For a 3x3 matrix, we can write the matrices \( A \) and \( B \) as:
\[ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \]
\[ B = \begin{bmatrix} 3^{0} a_{11} & 3^{1} a_{12} & 3^{2} a_{13} \\ 3^{1} a_{21} & 3^{2} a_{22} & 3^{3} a_{23} \\ 3^{2} a_{31} & 3^{3} a_{32} & 3^{4} a_{33} \end{bmatrix} \]

3. **Factoring out constants from columns**:
We can factor out powers of 3 from the columns of matrix \( B \):
- From the first column, we take out \( 3^0 \) (which is 1).
- From the second column, we take out \( 3^1 \).
- From the third column, we take out \( 3^2 \).

This gives us:
\[ B = 3^{0 + 1 + 2} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = 3^3 A \]

4. **Calculating the determinant**:
The determinant of matrix \( B \) can be expressed in terms of the determinant of matrix \( A \):
\[ |B| = |3^3 A| = 3^3 |A| \]
Thus,
\[ |B| = 27 |A| \]

5. **Using the given determinant of \( B \)**:
We know that \( |B| = 81 \). Therefore, we can set up the equation:
\[ 27 |A| = 81 \]

6. **Solving for \( |A| \)**:
Dividing both sides by 27 gives:
\[ |A| = \frac{81}{27} = 3 \]

### Final Answer:
The determinant of matrix \( A \) is:
\[ |A| = 3 \]