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 find the determinant of the matrix \( B \) given that the determinant of matrix \( A \) is \( 2 \), we will follow these steps:

Given:
\[ A = \begin{bmatrix} a & b & c \\ p & q & r \\ x & y & z \end{bmatrix} \]
and
\[ \text{det}(A) = 2 \]
We need to find:
\[ B = \begin{bmatrix} 4x & 2a & -p \\ 4y & 2b & -q \\ 4z & 2c & -r \end{bmatrix} \]

### Step 1: Factor out constants from the determinant
Using the property of determinants that allows us to factor out constants from rows or columns, we can rewrite the determinant of \( B \):
\[ \text{det}(B) = \text{det}\left(\begin{bmatrix} 4x & 2a & -p \\ 4y & 2b & -q \\ 4z & 2c & -r \end{bmatrix}\right) \]
We can factor out \( 4 \) from the first column, \( 2 \) from the second column, and \( -1 \) from the third column:
\[ \text{det}(B) = 4 \cdot 2 \cdot (-1) \cdot \text{det}\left(\begin{bmatrix} x & a & p \\ y & b & q \\ z & c & r \end{bmatrix}\right) \]
This simplifies to:
\[ \text{det}(B) = -8 \cdot \text{det}\left(\begin{bmatrix} x & a & p \\ y & b & q \\ z & c & r \end{bmatrix}\right) \]

### Step 2: Relate the determinant of \( B \) to \( A \)
Notice that the matrix inside the determinant is a rearrangement of the columns of \( A \). Specifically, we can express it as:
\[ \text{det}\left(\begin{bmatrix} x & a & p \\ y & b & q \\ z & c & r \end{bmatrix}\right) = \text{det}(A) \]
However, we will need to account for the column swaps.

### Step 3: Interchange columns
1. Interchange column 1 and column 2:
\[ \text{det}\left(\begin{bmatrix} x & a & p \\ y & b & q \\ z & c & r \end{bmatrix}\right) = -\text{det}\left(\begin{bmatrix} a & x & p \\ b & y & q \\ c & z & r \end{bmatrix}\right) \]
2. Interchange column 2 and column 3:
\[ \text{det}\left(\begin{bmatrix} a & x & p \\ b & y & q \\ c & z & r \end{bmatrix}\right) = -\text{det}(A) \]

### Step 4: Final expression for \( \text{det}(B) \)
Combining these results, we have:
\[ \text{det}(B) = -8 \cdot (-\text{det}(A)) = 8 \cdot \text{det}(A) \]
Substituting \( \text{det}(A) = 2 \):
\[ \text{det}(B) = 8 \cdot 2 = 16 \]

### Step 5: Apply the negative sign from the earlier factorization
Thus, we have:
\[ \text{det}(B) = -16 \]

### Conclusion
The final result is:
\[ \text{det}(B) = -16 \]