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

Options

A. A
B. A'
C. 3A
D. 3A'

Answer

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

### Step 1: Find the Cofactor Matrix
The cofactor matrix \( C \) is found by calculating the cofactor \( C_{ij} \) for each element of the matrix \( A \). The cofactor is given by:
\[ C_{ij} = (-1)^{i+j} \cdot \text{det}(M_{ij}) \]
where \( M_{ij} \) is the minor matrix obtained by deleting the \( i \)-th row and \( j \)-th column from \( A \).

#### Calculation of Cofactors:
1. **Cofactor \( C_{11} \)**:
\[ M_{11} = \begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix}, \quad \text{det}(M_{11}) = (1)(1) - (-2)(-2) = 1 - 4 = -3 \implies C_{11} = (-1)^{1+1} \cdot (-3) = -3 \]

2. **Cofactor \( C_{12} \)**:
\[ M_{12} = \begin{bmatrix} 2 & -2 \\ 2 & 1 \end{bmatrix}, \quad \text{det}(M_{12}) = (2)(1) - (-2)(2) = 2 + 4 = 6 \implies C_{12} = (-1)^{1+2} \cdot 6 = -6 \]

3. **Cofactor \( C_{13} \)**:
\[ M_{13} = \begin{bmatrix} 2 & 1 \\ 2 & -2 \end{bmatrix}, \quad \text{det}(M_{13}) = (2)(-2) - (1)(2) = -4 - 2 = -6 \implies C_{13} = (-1)^{1+3} \cdot (-6) = -6 \]

4. **Cofactor \( C_{21} \)**:
\[ M_{21} = \begin{bmatrix} -2 & -2 \\ -2 & 1 \end{bmatrix}, \quad \text{det}(M_{21}) = (-2)(1) - (-2)(-2) = -2 - 4 = -6 \implies C_{21} = (-1)^{2+1} \cdot (-6) = 6 \]

5. **Cofactor \( C_{22} \)**:
\[ M_{22} = \begin{bmatrix} -1 & -2 \\ 2 & 1 \end{bmatrix}, \quad \text{det}(M_{22}) = (-1)(1) - (-2)(2) = -1 + 4 = 3 \implies C_{22} = (-1)^{2+2} \cdot 3 = 3 \]

6. **Cofactor \( C_{23} \)**:
\[ M_{23} = \begin{bmatrix} -1 & -2 \\ 2 & -2 \end{bmatrix}, \quad \text{det}(M_{23}) = (-1)(-2) - (-2)(2) = 2 + 4 = 6 \implies C_{23} = (-1)^{2+3} \cdot 6 = -6 \]

7. **Cofactor \( C_{31} \)**:
\[ M_{31} = \begin{bmatrix} -2 & -2 \\ 1 & -2 \end{bmatrix}, \quad \text{det}(M_{31}) = (-2)(-2) - (-2)(1) = 4 + 2 = 6 \implies C_{31} = (-1)^{3+1} \cdot 6 = 6 \]

8. **Cofactor \( C_{32} \)**:
\[ M_{32} = \begin{bmatrix} -1 & -2 \\ 2 & -2 \end{bmatrix}, \quad \text{det}(M_{32}) = (-1)(-2) - (-2)(2) = 2 + 4 = 6 \implies C_{32} = (-1)^{3+2} \cdot 6 = -6 \]

9. **Cofactor \( C_{33} \)**:
\[ M_{33} = \begin{bmatrix} -1 & -2 \\ 2 & 1 \end{bmatrix}, \quad \text{det}(M_{33}) = (-1)(1) - (-2)(2) = -1 + 4 = 3 \implies C_{33} = (-1)^{3+3} \cdot 3 = 3 \]

#### Constructing the Cofactor Matrix:
\[ C = \begin{bmatrix} -3 & -6 & -6 \\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{bmatrix} \]

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

### Final Answer
Thus, the adjoint of matrix \( A \) is:
\[ \text{adj}(A) = \begin{bmatrix} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix} \]

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