Let `A= [[a,b,c],[p,q,r],[x,y,z]]` and suppose then det (A) = 2, then det (B) equals, where `B = [[4x,2a,-p],[4y, 2b, -q],[4z, 2c, -r]]`

Options

A. -2
B. -8
C. -16
D. 8

Explanation

To solve the problem, we need to find the determinant of the matrix \( B \) given that the determinant of matrix \( A \) is 2.

### Step-by-step Solution:

1. **Identify the matrices**:
We have:
\[ A = \begin{bmatrix} a & b & c \\ p & q & r \\ x & y & z \end{bmatrix} \]
and
\[ B = \begin{bmatrix} 4x & 2a & -p \\ 4y & 2b & -q \\ 4z & 2c & -r \end{bmatrix} \]

2. **Factor out constants from the columns of matrix \( B \)**:
- From the first column, we can factor out \( 4 \).
- From the second column, we can factor out \( 2 \).
- The third column has a factor of \( -1 \) (due to the negative signs).

Thus, we can rewrite the determinant of \( B \) as:
\[ \text{det}(B) = 4 \cdot 2 \cdot (-1) \cdot \text{det}\left(\begin{bmatrix} x & a & -p \\ y & b & -q \\ z & c & -r \end{bmatrix}\right) \]
This simplifies to:
\[ \text{det}(B) = -8 \cdot \text{det}\left(\begin{bmatrix} x & a & -p \\ y & b & -q \\ z & c & -r \end{bmatrix}\right) \]

3. **Relate the determinant of the modified matrix to \( A \)**:
The matrix we have now can be transformed back to a form similar to \( A \) by rearranging the columns. The determinant of \( A \) is given as \( 2 \).

We can express:
\[ \text{det}\left(\begin{bmatrix} x & a & -p \\ y & b & -q \\ z & c & -r \end{bmatrix}\right) = -\text{det}(A) \]
because we have interchanged columns, which introduces a negative sign.

4. **Substituting the determinant of \( A \)**:
Since \( \text{det}(A) = 2 \), we have:
\[ \text{det}\left(\begin{bmatrix} x & a & -p \\ y & b & -q \\ z & c & -r \end{bmatrix}\right) = -2 \]

5. **Final calculation**:
Now substituting back into the determinant of \( B \):
\[ \text{det}(B) = -8 \cdot (-2) = 16 \]

### Conclusion:
Thus, the determinant of matrix \( B \) is:
\[ \text{det}(B) = 16 \]