If `A=[[a_(11), a_(12), a_(13)], [a_(21), a_(22), a_(23)], [a_(31), a_(32), a_(33)]] and a_(ij)` denote the cofactor of element `A_(ij)`, then `a_(11)A_(11)+a_(12)A_(12)+a_(13)A_(13)=`

Options

A. `0`
B. `|A|`
C. `-|A|`
D. `2|A|`

Answer

To solve the problem, we need to find the value of the expression \( a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} \), where \( a_{ij} \) denotes the cofactor of the element \( A_{ij} \) of the matrix \( A \).

### Step-by-Step Solution:

1. **Understanding Cofactors**:
The cofactor \( a_{ij} \) of an element \( A_{ij} \) in a matrix is given by:
\[ a_{ij} = (-1)^{i+j} \cdot m_{ij} \]
where \( m_{ij} \) is the minor of the element \( A_{ij} \).

2. **Identify the Matrix**:
Given the matrix:
\[ A = \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} \]

3. **Calculate the Cofactors**:
- For \( a_{11} \):
\[ a_{11} = (-1)^{1+1} \cdot m_{11} = m_{11} \]
where \( m_{11} = \begin{vmatrix} A_{22} & A_{23} \\ A_{32} & A_{33} \end{vmatrix} = A_{22}A_{33} - A_{23}A_{32} \] - For \( a_{12} \):
\[ a_{12} = (-1)^{1+2} \cdot m_{12} = -m_{12} \]
where \( m_{12} = \begin{vmatrix} A_{21} & A_{23} \\ A_{31} & A_{33} \end{vmatrix} = A_{21}A_{33} - A_{23}A_{31} \] - For \( a_{13} \):
\[ a_{13} = (-1)^{1+3} \cdot m_{13} = m_{13} \]
where \( m_{13} = \begin{vmatrix} A_{21} & A_{22} \\ A_{31} & A_{32} \end{vmatrix} = A_{21}A_{32} - A_{22}A_{31} \] 4. **Substituting the Cofactors**: Now substitute the cofactors into the expression: \[ a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = m_{11}A_{11} - m_{12}A_{12} + m_{13}A_{13} \] 5. **Expanding the Expression**: Substitute the values of \( m_{11}, m_{12}, \) and \( m_{13} \):
\[ = (A_{22}A_{33} - A_{23}A_{32})A_{11} - (A_{21}A_{33} - A_{23}A_{31})A_{12} + (A_{21}A_{32} - A_{22}A_{31})A_{13} \]

6. **Recognizing the Determinant**:
The expression \( a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} \) represents the determinant of matrix \( A \):
\[ = \text{det}(A) \]

### Conclusion:
Thus, the final result is:
\[ a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = \text{det}(A) \]