`A=[[1, 0, 2], [-1, 1, -2], [0, 2, 1]] and adjA=[[5, x, -2], [1, 1, 0], [-2, -2, y]]`, then (x, y)=

Options

A. `(4, -1)`
B. `(-4, 1)`
C. `(-4, -10)`
D. `(4, 1)`

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) in the adjoint matrix \( \text{adj} A \) given the matrix \( A \).

### Step-by-Step Solution:

1. **Identify the Matrices**:
We have:
\[ A = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix} \]
and
\[ \text{adj} A = \begin{bmatrix} 5 & x & -2 \\ 1 & 1 & 0 \\ -2 & -2 & y \end{bmatrix} \]

2. **Find the Cofactor Matrix**:
The adjoint of a matrix is the transpose of its cofactor matrix. We need to find the cofactors corresponding to the positions of \( x \) and \( y \).

3. **Calculate \( C_{21} \) (Cofactor for \( x \))**:
The cofactor \( C_{21} \) is calculated by removing the second row and first column from \( A \):
\[ C_{21} = (-1)^{2+1} \cdot D_{21} \]
where \( D_{21} \) is the determinant of the matrix formed by the remaining elements:
\[ D_{21} = \begin{vmatrix} 0 & 2 \\ 2 & 1 \end{vmatrix} = (0 \cdot 1) - (2 \cdot 2) = -4 \]
Thus,
\[ C_{21} = -(-4) = 4 \]
Therefore, \( x = 4 \).

4. **Calculate \( C_{33} \) (Cofactor for \( y \))**:
The cofactor \( C_{33} \) is calculated by removing the third row and third column from \( A \):
\[ C_{33} = (-1)^{3+3} \cdot D_{33} \]
where \( D_{33} \) is the determinant of the matrix formed by the remaining elements:
\[ D_{33} = \begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} = (1 \cdot 1) - (0 \cdot -1) = 1 \]
Thus,
\[ C_{33} = 1 \]
Therefore, \( y = 1 \).

5. **Final Result**:
We have found \( x \) and \( y \):
\[ (x, y) = (4, 1) \]

### Conclusion:
The values of \( x \) and \( y \) are:
\[ \boxed{(4, 1)} \]