If `A` is a `3x3` matrix and `det (3A) = k {det(A)} , k` is equal to

Options

A. 9
B. 6
C. 1
D. 27

Answer

To solve the problem, we need to find the value of \( k \) such that \( \text{det}(3A) = k \cdot \text{det}(A) \) for a \( 3 \times 3 \) matrix \( A \).

### Step-by-Step Solution:

1. **Understanding the Determinant Property**:
For any square matrix \( X \) of order \( n \), the property of determinants states:
\[ \text{det}(kX) = k^n \cdot \text{det}(X) \]
where \( k \) is a scalar and \( n \) is the order of the matrix.

2. **Identify the Order of Matrix**:
Here, \( A \) is a \( 3 \times 3 \) matrix, so \( n = 3 \).

3. **Apply the Determinant Property**:
We need to find \( \text{det}(3A) \). According to the property:
\[ \text{det}(3A) = 3^3 \cdot \text{det}(A) \]

4. **Calculate \( 3^3 \)**:
Now, calculate \( 3^3 \):
\[ 3^3 = 27 \]

5. **Relate to the Given Equation**:
From the property, we have:
\[ \text{det}(3A) = 27 \cdot \text{det}(A) \]
This means that \( k \) in the equation \( \text{det}(3A) = k \cdot \text{det}(A) \) is equal to 27.

6. **Final Answer**:
Thus, the value of \( k \) is:
\[ k = 27 \]

### Summary:
The value of \( k \) is 27.