This problem involves the properties of determinants, specifically how scalar multiplication of rows or columns affects the determinant, and the relationship between the determinant of a matrix and its transpose. The key is to carefully apply the given rule for $b_{ij}$ and then factor out common terms from the rows and columns of the determinant of $B$.

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1. Key Concepts and Formulas

• Determinant of a Matrix: For a square matrix $M$, its determinant, denoted as $|M|$ or $\det(M)$, is a scalar value computed from its elements.
• Scalar Multiplication of a Row/Column: If a single row or column of a matrix $M$ is multiplied by a scalar $k$, the determinant of the new matrix is $k \cdot \det(M)$. This implies that if all elements of a row/column have a common factor $k$, this factor can be taken out of the determinant, multiplying the determinant by $k$.
• Determinant of a Scalar Multiple of a Matrix: If $M$ is an $n \times n$ matrix and $k$ is a scalar, then $\det(kM) = k^n \det(M)$. (This is a consequence of applying the previous rule to each of the $n$ rows).
• Determinant of a Transpose: For any square matrix $M$, the determinant of its transpose $M^T$ is equal to the determinant of $M$, i.e., $\det(M^T) = \det(M)$.

2. Step-by-Step Working with Explanations

Step 1: Express the elements of matrix B in terms of A. The problem states that $b_{ij} = (3)^{(i+j-2)} a_{ji}$ for a $3 \times 3$ matrix. We can write the exponent $(i+j-2)$ as $(i-1) + (j-1)$. So, $b_{ij} = 3^{(i-1)} \cdot 3^{(j-1)} a_{ji}$.

Let's write out the elements of matrix B explicitly for a $3 \times 3$ matrix: The powers of 3 will be:

• For $i=1$: $3^{1-1} = 3^0 = 1$
• For $i=2$: $3^{2-1} = 3^1 = 3$
• For $i=3$: $3^{3-1} = 3^2 = 9$

Similarly for $j$:

• For $j=1$: $3^{1-1} = 3^0 = 1$
• For $j=2$: $3^{2-1} = 3^1 = 3$
• For $j=3$: $3^{3-1} = 3^2 = 9$

Now, let's construct matrix B:

b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{pmatrix}$$ Substitute $b_{ij} = 3^{(i-1)} 3^{(j-1)} a_{ji}$: * $b_{11} = 3^0 \cdot 3^0 a_{11} = a_{11}$ * $b_{12} = 3^0 \cdot 3^1 a_{21} = 3a_{21}$ * $b_{13} = 3^0 \cdot 3^2 a_{31} = 9a_{31}$ * $b_{21} = 3^1 \cdot 3^0 a_{12} = 3a_{12}$ * $b_{22} = 3^1 \cdot 3^1 a_{22} = 9a_{22}$ * $b_{23} = 3^1 \cdot 3^2 a_{32} = 27a_{32}$ * $b_{31} = 3^2 \cdot 3^0 a_{13} = 9a_{13}$ * $b_{32} = 3^2 \cdot 3^1 a_{23} = 27a_{23}$ * $b_{33} = 3^2 \cdot 3^2 a_{33} = 81a_{33}$ So, matrix B is: $$B = \begin{pmatrix} a_{11} & 3a_{21} & 9a_{31} \\ 3a_{12} & 9a_{22} & 27a_{32} \\ 9a_{13} & 27a_{23} & 81a_{33} \end{pmatrix}$$ **Step 2: Calculate $\det(B)$ by factoring out scalars from rows.** We start with the determinant of B: $$\det(B) = \begin{vmatrix} a_{11} & 3a_{21} & 9a_{31} \\ 3a_{12} & 9a_{22} & 27a_{32} \\ 9a_{13} & 27a_{23} & 81a_{33} \end{vmatrix}$$ **Explanation:** We apply the property that a common factor in any row can be factored out. * From Row 1, we can factor out $3^0 = 1$. * From Row 2, we can factor out $3^1 = 3$. * From Row 3, we can factor out $3^2 = 9$. Factoring these out, we get: $$\det(B) = (3^0 \cdot 3^1 \cdot 3^2) \begin{vmatrix} a_{11} & 3a_{21} & 9a_{31} \\ a_{12} & 3a_{22} & 9a_{32} \\ a_{13} & 3a_{23} & 9a_{33} \end{vmatrix}$$ The product of the factors is $3^{0+1+2} = 3^3 = 27$. $$\det(B) = 27 \begin{vmatrix} a_{11} & 3a_{21} & 9a_{31} \\ a_{12} & 3a_{22} & 9a_{32} \\ a_{13} & 3a_{23} & 9a_{33} \end{vmatrix}$$ **Step 3: Factor out scalars from columns of the resulting determinant.** Now, we consider the remaining determinant: $$\begin{vmatrix} a_{11} & 3a_{21} & 9a_{31} \\ a_{12} & 3a_{22} & 9a_{32} \\ a_{13} & 3a_{23} & 9a_{33} \end{vmatrix}$$ **Explanation:** Similarly, a common factor in any column can be factored out. * From Column 1, there is no common factor other than $3^0 = 1$. * From Column 2, we can factor out $3^1 = 3$. * From Column 3, we can factor out $3^2 = 9$. Factoring these out, we get: $$\det(B) = 27 \cdot (3^0 \cdot 3^1 \cdot 3^2) \begin{vmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{vmatrix}$$ The product of these new factors is $3^{0+1+2} = 3^3 = 27$. So, $\det(B) = 27 \cdot 27 \begin{vmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{vmatrix}$ $\det(B) = 3^3 \cdot 3^3 \begin{vmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{vmatrix} = 3^6 \begin{vmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{vmatrix}$ **Step 4: Relate the resulting determinant to $\det(A)$.** The determinant on the right side is: $$\begin{vmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{vmatrix}$$ **Explanation:** This matrix is the transpose of matrix A ($A^T$). Recall that $(A^T)_{ij} = A_{ji}$. So, $\begin{vmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{vmatrix} = \det(A^T)$. Using the property $\det(A^T) = \det(A)$, we can write: $\det(B) = 3^6 \det(A)$. **Step 5: Solve for $\det(A)$.** We are given that $\det(B) = 81$. Substitute this into our derived equation: $81 = 3^6 \det(A)$. We know that $81 = 3^4$. So, $3^4 = 3^6 \det(A)$. To find $\det(A)$, divide both sides by $3^6$: $\det(A) = \frac{3^4}{3^6} = 3^{4-6} = 3^{-2}$. $\det(A) = \frac{1}{3^2} = \frac{1}{9}$. **Important Note:** The mathematical derivation based on the provided problem statement $b_{ij} = (3)^{(i+j-2)} a_{ji}$ consistently leads to $\det(A) = \frac{1}{9}$. If the intended answer is (A) 3, it would imply a different relationship, such as $b_{ij} = 3^{i-1} a_{ji}$ or $b_{ij} = 3^{j-1} a_{ji}$, in which case the total factor would be $3^3$ instead of $3^6$. However, strictly following the given problem statement, the answer is $1/9$. The final answer is $\boxed{\frac{1}{9}}$. **3. Tips and Common Mistakes** * **Careful with Exponents:** Ensure the exponent $(i+j-2)$ is correctly applied to the base $3$. It's $3$ raised to the power of $(i+j-2)$, not $3 \times (i+j-2)$. * **Order of Indices:** Pay close attention to $a_{ji}$ versus $a_{ij}$. This indicates the involvement of the transpose of matrix A. Forgetting this will lead to an incorrect result. * **Systematic Factoring:** When factoring scalars from a determinant, it's crucial to do it systematically (e.g., all rows first, then all columns, or vice-versa). Ensure all common factors are extracted. Each row/column scaling affects the determinant multiplicatively. * **Determinant Properties:** Remember $\det(A^T) = \det(A)$ and $\det(kA) = k^n \det(A)$ for an $n \times n$ matrix. The latter is a special case of factoring from all rows (or columns). **4. Summary / Key Takeaway** This problem demonstrates the importance of meticulously applying determinant properties when matrix elements are defined by a complex rule. By expressing each element $b_{ij}$ and then systematically factoring out common terms from rows and columns, we can relate $\det(B)$ to $\det(A^T)$, and subsequently to $\det(A)$. The product of all factored scalars becomes the coefficient multiplying $\det(A)$. For a $3 \times 3$ matrix with elements $b_{ij} = 3^{(i-1)} 3^{(j-1)} a_{ji}$, the total scaling factor is $3^3 \cdot 3^3 = 3^6$.