The adjoint of the matrix `A=[{:(1,0,2),(2,1,0),(0,3,1):}]`is

Options

A. `[{:(-1,6,2),(-2,1,-4),(6,3,1):}]`
B. `[{:(1,6,-2),(-2,1,4),(6,-3,1):}]`
C. `[{:(6,1,2),(4,-1,2),(6,3,-1):}]`
D. `[{:(-6,1,2),(4,-2,1),(3,1,-6):}]`

Explanation

To find the adjoint of the matrix \( A = \begin{pmatrix} 1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 3 & 1 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Cofactors
The adjoint of a matrix is the transpose of the cofactor matrix. We will calculate the cofactor for each element of the matrix.

#### Cofactor Calculation
1. **Cofactor of \( a_{11} = 1 \)**:
- Remove the first row and first column:
\[ \text{Minor} = \begin{vmatrix} 1 & 0 \\ 3 & 1 \end{vmatrix} = (1)(1) - (0)(3) = 1 \]
- Cofactor \( C_{11} = (-1)^{1+1} \cdot 1 = 1 \)

2. **Cofactor of \( a_{12} = 0 \)**:
- Remove the first row and second column:
\[ \text{Minor} = \begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} = (2)(1) - (0)(0) = 2 \]
- Cofactor \( C_{12} = (-1)^{1+2} \cdot 2 = -2 \)

3. **Cofactor of \( a_{13} = 2 \)**:
- Remove the first row and third column:
\[ \text{Minor} = \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix} = (2)(3) - (1)(0) = 6 \]
- Cofactor \( C_{13} = (-1)^{1+3} \cdot 6 = 6 \)

4. **Cofactor of \( a_{21} = 2 \)**:
- Remove the second row and first column:
\[ \text{Minor} = \begin{vmatrix} 0 & 2 \\ 3 & 1 \end{vmatrix} = (0)(1) - (2)(3) = -6 \]
- Cofactor \( C_{21} = (-1)^{2+1} \cdot (-6) = 6 \)

5. **Cofactor of \( a_{22} = 1 \)**:
- Remove the second row and second column:
\[ \text{Minor} = \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = (1)(1) - (2)(0) = 1 \]
- Cofactor \( C_{22} = (-1)^{2+2} \cdot 1 = 1 \)

6. **Cofactor of \( a_{23} = 0 \)**:
- Remove the second row and third column:
\[ \text{Minor} = \begin{vmatrix} 1 & 0 \\ 0 & 3 \end{vmatrix} = (1)(3) - (0)(0) = 3 \]
- Cofactor \( C_{23} = (-1)^{2+3} \cdot 3 = -3 \)

7. **Cofactor of \( a_{31} = 0 \)**:
- Remove the third row and first column:
\[ \text{Minor} = \begin{vmatrix} 0 & 2 \\ 1 & 0 \end{vmatrix} = (0)(0) - (2)(1) = -2 \]
- Cofactor \( C_{31} = (-1)^{3+1} \cdot (-2) = -2 \)

8. **Cofactor of \( a_{32} = 3 \)**:
- Remove the third row and second column:
\[ \text{Minor} = \begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} = (1)(0) - (2)(2) = -4 \]
- Cofactor \( C_{32} = (-1)^{3+2} \cdot (-4) = 4 \)

9. **Cofactor of \( a_{33} = 1 \)**:
- Remove the third row and third column:
\[ \text{Minor} = \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = (1)(1) - (0)(2) = 1 \]
- Cofactor \( C_{33} = (-1)^{3+3} \cdot 1 = 1 \)

### Step 2: Form the Cofactor Matrix
The cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} 1 & -2 & 6 \\ 6 & 1 & -3 \\ -2 & 4 & 1 \end{pmatrix} \]

### Step 3: Transpose the Cofactor Matrix
The adjoint of matrix \( A \) is the transpose of the cofactor matrix \( C \):
\[ \text{adj}(A) = C^T = \begin{pmatrix} 1 & 6 & -2 \\ -2 & 1 & 4 \\ 6 & -3 & 1 \end{pmatrix} \]

### Final Answer
The adjoint of the matrix \( A \) is:
\[ \text{adj}(A) = \begin{pmatrix} 1 & 6 & -2 \\ -2 & 1 & 4 \\ 6 & -3 & 1 \end{pmatrix} \]