Let `|A|=|a_(ij)|_(3xx3) ne0` Each element `a_(ij)` is multiplied by by `k^(i-j)` Let `|B|` the resulting determinant, where `k_1 |A|+k_2 |B| =a` then `k_1+k_2 =`

Options

A. `1`
B. `-1`
C. `0`
D. `2`

Explanation

To solve the problem, we need to analyze the transformation of the determinant \( |A| \) when each element \( a_{ij} \) is multiplied by \( k^{i-j} \). Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Understanding the Determinant \( |A| \)**:
Given a \( 3 \times 3 \) matrix \( A = [a_{ij}] \), where \( |A| \neq 0 \).

2. **Transforming the Matrix**:
Each element \( a_{ij} \) of the matrix \( A \) is multiplied by \( k^{i-j} \). This results in a new matrix \( B \) where:
\[ b_{ij} = a_{ij} \cdot k^{i-j} \]
Thus, the matrix \( B \) can be expressed as:
\[ B = \begin{bmatrix} a_{11} & k^{-1} a_{12} & k^{-2} a_{13} \\ k a_{21} & a_{22} & k^{-1} a_{23} \\ k^2 a_{31} & k a_{32} & a_{33} \end{bmatrix} \]

3. **Calculating the Determinant \( |B| \)**:
The determinant \( |B| \) can be calculated using the properties of determinants. Specifically, when we factor out \( k \) from each row:
- From the first row, we factor out \( k^0 \) (no factor),
- From the second row, we factor out \( k^1 \),
- From the third row, we factor out \( k^2 \).

Therefore, the determinant \( |B| \) can be expressed as:
\[ |B| = k^0 \cdot k^1 \cdot k^2 \cdot |A| = k^3 |A| \]

4. **Setting Up the Equation**:
According to the problem, we have:
\[ k_1 |A| + k_2 |B| = 0 \]
Substituting \( |B| \) from the previous step:
\[ k_1 |A| + k_2 (k^3 |A|) = 0 \]

5. **Factoring Out \( |A| \)**:
Since \( |A| \neq 0 \), we can divide through by \( |A| \):
\[ k_1 + k_2 k^3 = 0 \]

6. **Finding \( k_1 + k_2 \)**:
Rearranging gives:
\[ k_1 = -k_2 k^3 \]
Now, adding \( k_1 + k_2 \):
\[ k_1 + k_2 = -k_2 k^3 + k_2 = k_2(1 - k^3) \]

7. **Conclusion**:
Since we need to find \( k_1 + k_2 \) in terms of \( k_2 \):
If \( k_2 \) is not zero, we can conclude that:
\[ k_1 + k_2 = 0 \quad \text{(if } k_2 \text{ is chosen such that } 1 - k^3 = 0\text{)} \]

Thus, the final result is:
\[ \boxed{0} \]