If, in `D={:[(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))]:},` the co-factor of `a_(r)" is "A_(r),`then , `c_(1)A_(1)+c_(2)A_(2)+c_(3)A_(3)=`

Options

A. 0
B. `-|D|`
C. `|D|`
D. `|D|^(2)`

Answer

To solve the problem, we need to find the expression \( c_1 A_1 + c_2 A_2 + c_3 A_3 \), where \( A_1, A_2, A_3 \) are the cofactors corresponding to the elements \( a_1, a_2, a_3 \) of the matrix \( D \).

### Step-by-Step Solution:

1. **Identify the Cofactors**:
- The cofactor \( A_r \) of an element \( a_r \) in a matrix is given by:
\[ A_r = (-1)^{r+c} \cdot M_{rc} \]
where \( M_{rc} \) is the determinant of the submatrix formed by deleting the \( r \)-th row and \( c \)-th column.

2. **Calculate \( A_1 \)**:
- For \( A_1 \) (cofactor of \( a_1 \)):
\[ A_1 = (-1)^{1+1} \cdot \text{det}\begin{pmatrix} b_2 & c_2 \\ b_3 & c_3 \end{pmatrix} = \text{det}\begin{pmatrix} b_2 & c_2 \\ b_3 & c_3 \end{pmatrix} = b_2 c_3 - b_3 c_2 \]

3. **Calculate \( A_2 \)**:
- For \( A_2 \) (cofactor of \( a_2 \)):
\[ A_2 = (-1)^{2+1} \cdot \text{det}\begin{pmatrix} b_1 & c_1 \\ b_3 & c_3 \end{pmatrix} = -\text{det}\begin{pmatrix} b_1 & c_1 \\ b_3 & c_3 \end{pmatrix} = - (b_1 c_3 - b_3 c_1) = b_3 c_1 - b_1 c_3 \]

4. **Calculate \( A_3 \)**:
- For \( A_3 \) (cofactor of \( a_3 \)):
\[ A_3 = (-1)^{3+1} \cdot \text{det}\begin{pmatrix} b_1 & c_1 \\ b_2 & c_2 \end{pmatrix} = \text{det}\begin{pmatrix} b_1 & c_1 \\ b_2 & c_2 \end{pmatrix} = b_1 c_2 - b_2 c_1 \]

5. **Substitute into the Expression**:
- Now substitute \( A_1, A_2, A_3 \) into the expression \( c_1 A_1 + c_2 A_2 + c_3 A_3 \):
\[ c_1 A_1 + c_2 A_2 + c_3 A_3 = c_1 (b_2 c_3 - b_3 c_2) + c_2 (b_3 c_1 - b_1 c_3) + c_3 (b_1 c_2 - b_2 c_1) \]

6. **Expand the Expression**:
- Expanding gives:
\[ = c_1 b_2 c_3 - c_1 b_3 c_2 + c_2 b_3 c_1 - c_2 b_1 c_3 + c_3 b_1 c_2 - c_3 b_2 c_1 \]

7. **Combine Like Terms**:
- Notice that terms cancel out:
\[ = (c_1 b_2 c_3 - c_3 b_2 c_1) + (c_2 b_3 c_1 - c_1 b_3 c_2) + (c_3 b_1 c_2 - c_2 b_1 c_3) = 0 \]

### Final Result:
Thus, we conclude that:
\[ c_1 A_1 + c_2 A_2 + c_3 A_3 = 0 \]