suppose `D= |{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}| ` and `Dprime= |{:(a_(1)+pb_(1),,b_(1)+qc_(1),,c_(1)+ra_(1)),(a_(2)+pb_(2),,b_(2)+qc_(2),,c_(2)+ra_(2)),(a_(3)+pb_(3),,b_(3)+qc_(3),,c_(3)+ra_(3)):}| `. Then

Options

A. `D'=D`
B. `D'=D(1-pqr)`
C. `D=D(1+p+q+r)`
D. `D'=D(1+pqr)`

Explanation

To solve the problem, we will analyze the determinants \( D \) and \( D' \) and use properties of determinants to express \( D' \) in terms of \( D \).

### Step-by-Step Solution:

1. **Define the Determinants**:
\[ D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \]
\[ D' = \begin{vmatrix} a_1 + p b_1 & b_1 + q c_1 & c_1 + r a_1 \\ a_2 + p b_2 & b_2 + q c_2 & c_2 + r a_2 \\ a_3 + p b_3 & b_3 + q c_3 & c_3 + r a_3 \end{vmatrix} \]

2. **Apply the Property of Determinants**:
We can express \( D' \) as the sum of two determinants. Using the property that if one row of a determinant is expressed as a sum, we can separate it into two determinants:
\[ D' = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + \begin{vmatrix} p b_1 & q c_1 & r a_1 \\ p b_2 & q c_2 & r a_2 \\ p b_3 & q c_3 & r a_3 \end{vmatrix} \]
This gives us:
\[ D' = D + \begin{vmatrix} p b_1 & q c_1 & r a_1 \\ p b_2 & q c_2 & r a_2 \\ p b_3 & q c_3 & r a_3 \end{vmatrix} \]

3. **Factor Out Constants**:
We can factor out \( p \), \( q \), and \( r \) from the second determinant:
\[ D' = D + p \cdot q \cdot r \begin{vmatrix} b_1 & c_1 & a_1 \\ b_2 & c_2 & a_2 \\ b_3 & c_3 & a_3 \end{vmatrix} \]
Notice that the second determinant is actually \( D \) with columns rearranged, which does not change its value. Therefore:
\[ D' = D + pqr \cdot D \]

4. **Combine the Terms**:
We can combine the terms:
\[ D' = D(1 + pqr) \]

### Final Result:
Thus, we have established the relationship between \( D \) and \( D' \):
\[ D' = D(1 + pqr) \]