Let A be a square matrix of order `3 times 3`, then `abs(kA)` is equal to

Options

A. `kabs(A)`
B. `k^(2)abs(A)`
C. `k^(3)abs(A)`
D. `3kabs(A)`

Answer

To solve the problem, we need to find the determinant of the matrix \( kA \), where \( A \) is a square matrix of order \( 3 \times 3 \) and \( k \) is a scalar.

### Step-by-Step Solution:

1. **Understanding the Matrix**:
- Let \( A \) be a \( 3 \times 3 \) matrix. The determinant of a matrix is a scalar value that can be computed from its elements.

2. **Applying the Scalar Multiplication Rule**:
- When a matrix \( A \) is multiplied by a scalar \( k \), the determinant of the resulting matrix \( kA \) can be expressed using the following property:
\[ \text{det}(kA) = k^n \cdot \text{det}(A) \]
- Here, \( n \) is the order of the matrix. Since \( A \) is a \( 3 \times 3 \) matrix, \( n = 3 \).

3. **Substituting the Values**:
- We substitute \( n = 3 \) into the formula:
\[ \text{det}(kA) = k^3 \cdot \text{det}(A) \]

4. **Final Result**:
- Therefore, we conclude that:
\[ \text{det}(kA) = k^3 \cdot \text{det}(A) \]

### Conclusion:
The determinant of the matrix \( kA \) is equal to \( k^3 \cdot \text{det}(A) \).