If `A=[[1, -1, 2], [-2, 3, 5], [-2, 0, -1]]`, then adjA=

Options

A. `[[-3, -1, -11], [-12, 3, -9], [6, 2, 1]]`
B. `[[-3, -12, 6], [-1, 3, 2], [-11, -9, 1]]`
C. `[[-3, -1, -11], [12, 3, -9], [6, 2, 1]]`
D. `[[-3, -1, 11], [-12, 2, 9], [6, 2, 1]]`

Answer

To find the adjoint of the matrix \( A \), we will follow these steps:

Given:
\[ A = \begin{bmatrix} 1 & -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1 \end{bmatrix} \]

### Step 1: Find the Cofactor Matrix \( C \)

The cofactor matrix \( C \) is obtained by calculating the cofactors \( C_{ij} \) for each element of the matrix \( A \).

The cofactor \( C_{ij} \) is given by:
\[ C_{ij} = (-1)^{i+j} \cdot D_{ij} \]
where \( D_{ij} \) is the determinant of the submatrix obtained by deleting the \( i \)-th row and \( j \)-th column from \( A \).

#### Calculate each cofactor:

1. **Cofactor \( C_{11} \)**:
\[ D_{11} = \begin{vmatrix} 3 & 5 \\ 0 & -1 \end{vmatrix} = (3 \cdot -1) - (5 \cdot 0) = -3 \]
\[ C_{11} = (-1)^{1+1} \cdot D_{11} = 1 \cdot (-3) = -3 \]

2. **Cofactor \( C_{12} \)**:
\[ D_{12} = \begin{vmatrix} -2 & 5 \\ -2 & -1 \end{vmatrix} = (-2 \cdot -1) - (5 \cdot -2) = 2 + 10 = 12 \]
\[ C_{12} = (-1)^{1+2} \cdot D_{12} = -1 \cdot 12 = -12 \]

3. **Cofactor \( C_{13} \)**:
\[ D_{13} = \begin{vmatrix} -2 & 3 \\ -2 & 0 \end{vmatrix} = (-2 \cdot 0) - (3 \cdot -2) = 0 + 6 = 6 \]
\[ C_{13} = (-1)^{1+3} \cdot D_{13} = 1 \cdot 6 = 6 \]

4. **Cofactor \( C_{21} \)**:
\[ D_{21} = \begin{vmatrix} -1 & 2 \\ 0 & -1 \end{vmatrix} = (-1 \cdot -1) - (2 \cdot 0) = 1 \]
\[ C_{21} = (-1)^{2+1} \cdot D_{21} = -1 \cdot 1 = -1 \]

5. **Cofactor \( C_{22} \)**:
\[ D_{22} = \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} = (1 \cdot -1) - (2 \cdot -2) = -1 + 4 = 3 \]
\[ C_{22} = (-1)^{2+2} \cdot D_{22} = 1 \cdot 3 = 3 \]

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

7. **Cofactor \( C_{31} \)**:
\[ D_{31} = \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = (-1 \cdot 5) - (2 \cdot 3) = -5 - 6 = -11 \]
\[ C_{31} = (-1)^{3+1} \cdot D_{31} = 1 \cdot -11 = -11 \]

8. **Cofactor \( C_{32} \)**:
\[ D_{32} = \begin{vmatrix} 1 & 2 \\ -2 & 5 \end{vmatrix} = (1 \cdot 5) - (2 \cdot -2) = 5 + 4 = 9 \]
\[ C_{32} = (-1)^{3+2} \cdot D_{32} = -1 \cdot 9 = -9 \]

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

### Step 2: Construct the Cofactor Matrix \( C \)

Now we can construct the cofactor matrix \( C \):
\[ C = \begin{bmatrix} -3 & -12 & 6 \\ -1 & 3 & 2 \\ -11 & -9 & 1 \end{bmatrix} \]

### Step 3: Find the Adjoint Matrix \( \text{adj}(A) \)

The adjoint of matrix \( A \) is the transpose of the cofactor matrix \( C \):
\[ \text{adj}(A) = C^T = \begin{bmatrix} -3 & -1 & -11 \\ -12 & 3 & -9 \\ 6 & 2 & 1 \end{bmatrix} \]

### Final Answer

Thus, the adjoint of matrix \( A \) is:
\[ \text{adj}(A) = \begin{bmatrix} -3 & -1 & -11 \\ -12 & 3 & -9 \\ 6 & 2 & 1 \end{bmatrix} \]