If `A=[a_(ij)]` is a matrix of order `2xx2`, such that `|A|=-15` and `c_(ij)` represents co-factor of `a_(ij)`, then find `a_(21)c_(21)+a_(22)c_(22)`.

Answer

To solve the problem, we need to find the expression \( a_{21}c_{21} + a_{22}c_{22} \) given that the determinant of matrix \( A \) is \( |A| = -15 \).

### Step-by-step Solution:

1. **Define the Matrix \( A \)**:
Let \( A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \).

2. **Calculate the Determinant**:
The determinant of matrix \( A \) is given by:
\[ |A| = a_{11}a_{22} - a_{12}a_{21} \]
We know from the problem statement that:
\[ |A| = -15 \]
Therefore, we have:
\[ a_{11}a_{22} - a_{12}a_{21} = -15 \quad \text{(Equation 1)} \]

3. **Find the Cofactors**:
The cofactor \( c_{ij} \) of an element \( a_{ij} \) in a 2x2 matrix is calculated as follows:
- \( c_{21} = (-1)^{2+1} \cdot |A_{21}| = -a_{12} \) (where \( A_{21} \) is the minor of \( a_{21} \))
- \( c_{22} = (-1)^{2+2} \cdot |A_{22}| = a_{11} \) (where \( A_{22} \) is the minor of \( a_{22} \))

4. **Substitute the Cofactors**:
Now we substitute the cofactors into the expression:
\[ a_{21}c_{21} + a_{22}c_{22} = a_{21}(-a_{12}) + a_{22}(a_{11}) \]
Simplifying this, we get:
\[ = -a_{21}a_{12} + a_{22}a_{11} \]

5. **Relate to the Determinant**:
From Equation 1, we know:
\[ a_{11}a_{22} - a_{12}a_{21} = -15 \]
Thus, we can rewrite:
\[ -a_{21}a_{12} + a_{22}a_{11} = -15 \]

6. **Final Result**:
Therefore, we conclude that:
\[ a_{21}c_{21} + a_{22}c_{22} = -15 \]

### Final Answer:
\[ a_{21}c_{21} + a_{22}c_{22} = -15 \]