Find the adjoint of matrix `A = [(2,0,-1),(3,1,2),(-1,1,2)]`

Answer

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

### Step 1: Calculate the Minors

The first step in finding the adjoint is to calculate the minors of each element of the matrix.

1. **Minor \( M_{11} \)**: Remove the first row and first column:
\[ M_{11} = \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = (1)(2) - (2)(1) = 2 - 2 = 0 \]

2. **Minor \( M_{12} \)**: Remove the first row and second column:
\[ M_{12} = \begin{vmatrix} 3 & 2 \\ -1 & 2 \end{vmatrix} = (3)(2) - (2)(-1) = 6 + 2 = 8 \]

3. **Minor \( M_{13} \)**: Remove the first row and third column:
\[ M_{13} = \begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} = (3)(1) - (1)(-1) = 3 + 1 = 4 \]

4. **Minor \( M_{21} \)**: Remove the second row and first column:
\[ M_{21} = \begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix} = (0)(2) - (-1)(1) = 0 + 1 = 1 \]

5. **Minor \( M_{22} \)**: Remove the second row and second column:
\[ M_{22} = \begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} = (2)(2) - (-1)(-1) = 4 - 1 = 3 \]

6. **Minor \( M_{23} \)**: Remove the second row and third column:
\[ M_{23} = \begin{vmatrix} 2 & 0 \\ -1 & 1 \end{vmatrix} = (2)(1) - (0)(-1) = 2 - 0 = 2 \]

7. **Minor \( M_{31} \)**: Remove the third row and first column:
\[ M_{31} = \begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix} = (0)(2) - (-1)(1) = 0 + 1 = 1 \]

8. **Minor \( M_{32} \)**: Remove the third row and second column:
\[ M_{32} = \begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} = (2)(2) - (-1)(3) = 4 + 3 = 7 \]

9. **Minor \( M_{33} \)**: Remove the third row and third column:
\[ M_{33} = \begin{vmatrix} 2 & 0 \\ 3 & 1 \end{vmatrix} = (2)(1) - (0)(3) = 2 - 0 = 2 \]

### Step 2: Calculate the Cofactors

Next, we compute the cofactors by applying the sign changes based on the position of the elements:

\[ C_{ij} = (-1)^{i+j} M_{ij} \]

1. **Cofactor \( C_{11} \)**: \( C_{11} = M_{11} = 0 \)
2. **Cofactor \( C_{12} \)**: \( C_{12} = -M_{12} = -8 \)
3. **Cofactor \( C_{13} \)**: \( C_{13} = M_{13} = 4 \)
4. **Cofactor \( C_{21} \)**: \( C_{21} = -M_{21} = -1 \)
5. **Cofactor \( C_{22} \)**: \( C_{22} = M_{22} = 3 \)
6. **Cofactor \( C_{23} \)**: \( C_{23} = -M_{23} = -2 \)
7. **Cofactor \( C_{31} \)**: \( C_{31} = M_{31} = 1 \)
8. **Cofactor \( C_{32} \)**: \( C_{32} = -M_{32} = -7 \)
9. **Cofactor \( C_{33} \)**: \( C_{33} = M_{33} = 2 \)

### Step 3: Form the Cofactor Matrix

The cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} 0 & -8 & 4 \\ -1 & 3 & -2 \\ 1 & -7 & 2 \end{pmatrix} \]

### Step 4: Transpose the Cofactor Matrix

The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = C^T = \begin{pmatrix} 0 & -1 & 1 \\ -8 & 3 & -7 \\ 4 & -2 & 2 \end{pmatrix} \]

### Final Answer

Thus, the adjoint of the matrix \( A \) is:
\[ \text{adj}(A) = \begin{pmatrix} 0 & -1 & 1 \\ -8 & 3 & -7 \\ 4 & -2 & 2 \end{pmatrix} \]