For a `2^(nd)` order determinant `abs(A)=abs(a_(ij))`, find `a_(12)A_(11)+a_(22)A_(21)`

Options

A. `abs(A)`
B. 0
C. 1
D. `-abs(A)`

Answer

To solve the problem, we need to find the expression \( a_{12} A_{11} + a_{22} A_{21} \) for a \( 2^{nd} \) order determinant \( A \) defined by the elements \( a_{ij} \).

### Step-by-step Solution:

1. **Define the 2x2 Matrix**:
Let the matrix \( A \) be defined as:
\[ A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \]

2. **Calculate the Determinant**:
The determinant \( |A| \) of the matrix \( A \) is given by:
\[ |A| = a_{11} a_{22} - a_{12} a_{21} \]

3. **Find the Adjoint Elements**:
For a \( 2 \times 2 \) matrix, the adjoint elements are defined as:
\[ A_{11} = a_{22}, \quad A_{12} = -a_{12}, \quad A_{21} = -a_{21}, \quad A_{22} = a_{11} \]

4. **Substitute the Adjoint Elements**:
Now, we substitute \( A_{11} \) and \( A_{21} \) into the expression \( a_{12} A_{11} + a_{22} A_{21} \):
\[ a_{12} A_{11} + a_{22} A_{21} = a_{12} a_{22} + a_{22} (-a_{21}) = a_{12} a_{22} - a_{22} a_{21} \]

5. **Factor Out \( a_{22} \)**:
We can factor out \( a_{22} \):
\[ a_{12} a_{22} - a_{22} a_{21} = a_{22} (a_{12} - a_{21}) \]

6. **Conclusion**:
Thus, the expression \( a_{12} A_{11} + a_{22} A_{21} \) simplifies to:
\[ a_{22} (a_{12} - a_{21}) \]

### Final Result:
The final result is:
\[ a_{12} A_{11} + a_{22} A_{21} = a_{22} (a_{12} - a_{21}) \]