If `{:A=[(1,2,2),(2,3,0),(0,1,2)]and adjA=[(6,-2,-6),(-4,2,x),(y,-1,-1)]:}`,then x + y =

Options

A. 6
B. -1
C. 3
D. 1

Answer

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

Given:
\[ A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 3 & 0 \\ 0 & 1 & 2 \end{pmatrix} \]
\[ \text{adj} A = \begin{pmatrix} 6 & -2 & -6 \\ -4 & 2 & x \\ y & -1 & -1 \end{pmatrix} \]

### Step 1: Calculate the determinant of matrix \( A \)

To find the adjoint matrix, we first need to calculate the determinant of matrix \( A \).

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 3 & 0 \\ 1 & 2 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 0 \\ 0 & 2 \end{vmatrix} + 2 \cdot \begin{vmatrix} 2 & 3 \\ 0 & 1 \end{vmatrix} \]

Calculating the minors:
- \( \begin{vmatrix} 3 & 0 \\ 1 & 2 \end{vmatrix} = (3 \cdot 2) - (0 \cdot 1) = 6 \)
- \( \begin{vmatrix} 2 & 0 \\ 0 & 2 \end{vmatrix} = (2 \cdot 2) - (0 \cdot 0) = 4 \)
- \( \begin{vmatrix} 2 & 3 \\ 0 & 1 \end{vmatrix} = (2 \cdot 1) - (3 \cdot 0) = 2 \)

Now substituting back into the determinant formula:
\[ \text{det}(A) = 1 \cdot 6 - 2 \cdot 4 + 2 \cdot 2 = 6 - 8 + 4 = 2 \]

### Step 2: Find the cofactor matrix of \( A \)

Next, we calculate the cofactor matrix \( C \) of \( A \).

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = (-1)^{1+1} \cdot \begin{vmatrix} 3 & 0 \\ 1 & 2 \end{vmatrix} = 6 \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = (-1)^{1+2} \cdot \begin{vmatrix} 2 & 0 \\ 0 & 2 \end{vmatrix} = -4 \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = (-1)^{1+3} \cdot \begin{vmatrix} 2 & 3 \\ 0 & 1 \end{vmatrix} = 2 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = (-1)^{2+1} \cdot \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} = -2 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = (-1)^{2+2} \cdot \begin{vmatrix} 1 & 2 \\ 0 & 2 \end{vmatrix} = 2 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = (-1)^{2+3} \cdot \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = -1 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = (-1)^{3+1} \cdot \begin{vmatrix} 2 & 2 \\ 3 & 0 \end{vmatrix} = -6 \]

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = (-1)^{3+2} \cdot \begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} = 4 \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = (-1)^{3+3} \cdot \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = -1 \]

Thus, the cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} 6 & -4 & 2 \\ -2 & 2 & -1 \\ -6 & 4 & -1 \end{pmatrix} \]

### Step 3: Find the adjoint of \( A \)

The adjoint of \( A \) is the transpose of the cofactor matrix:
\[ \text{adj} A = C^T = \begin{pmatrix} 6 & -2 & -6 \\ -4 & 2 & 4 \\ 2 & -1 & -1 \end{pmatrix} \]

### Step 4: Compare with the given adjoint matrix

Now we compare the calculated adjoint matrix with the given adjoint matrix:
\[ \text{adj} A = \begin{pmatrix} 6 & -2 & -6 \\ -4 & 2 & x \\ y & -1 & -1 \end{pmatrix} \]

From the comparison, we can see:
- \( x = 4 \)
- \( y = 2 \)

### Step 5: Calculate \( x + y \)

Now we can find \( x + y \):
\[ x + y = 4 + 2 = 6 \]

### Final Answer
\[ \boxed{6} \]