If `A=[(-1,-2,-2),(2,1,-2),(2,-2,1)]` the adj. A=

Options

A. A
B. `A^(T)`
C. `3A`
D. `3A^(T)`

Answer

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

1. **Define the matrix \( A \)**:
\[ A = \begin{pmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix} \]

2. **Find the cofactor matrix of \( A \)**:
The cofactor matrix is obtained by calculating the cofactor for each element of the matrix \( A \). The cofactor \( C_{ij} \) is given by:
\[ C_{ij} = (-1)^{i+j} \cdot M_{ij} \]
where \( M_{ij} \) is the determinant of the submatrix obtained by deleting the \( i \)-th row and \( j \)-th column.

- **Cofactor \( C_{11} \)**:
\[ M_{11} = \begin{vmatrix} 1 & -2 \\ -2 & 1 \end{vmatrix} = (1)(1) - (-2)(-2) = 1 - 4 = -3 \]
\[ C_{11} = (-1)^{1+1} \cdot (-3) = -3 \]

- **Cofactor \( C_{12} \)**:
\[ M_{12} = \begin{vmatrix} 2 & -2 \\ 2 & 1 \end{vmatrix} = (2)(1) - (-2)(2) = 2 + 4 = 6 \]
\[ C_{12} = (-1)^{1+2} \cdot 6 = -6 \]

- **Cofactor \( C_{13} \)**:
\[ M_{13} = \begin{vmatrix} 2 & 1 \\ 2 & -2 \end{vmatrix} = (2)(-2) - (1)(2) = -4 - 2 = -6 \]
\[ C_{13} = (-1)^{1+3} \cdot (-6) = -6 \]

- **Cofactor \( C_{21} \)**:
\[ M_{21} = \begin{vmatrix} -2 & -2 \\ -2 & 1 \end{vmatrix} = (-2)(1) - (-2)(-2) = -2 - 4 = -6 \]
\[ C_{21} = (-1)^{2+1} \cdot (-6) = 6 \]

- **Cofactor \( C_{22} \)**:
\[ M_{22} = \begin{vmatrix} -1 & -2 \\ 2 & -2 \end{vmatrix} = (-1)(-2) - (-2)(2) = 2 + 4 = 6 \]
\[ C_{22} = (-1)^{2+2} \cdot 6 = 6 \]

- **Cofactor \( C_{23} \)**:
\[ M_{23} = \begin{vmatrix} -1 & -2 \\ 2 & 1 \end{vmatrix} = (-1)(1) - (-2)(2) = -1 + 4 = 3 \]
\[ C_{23} = (-1)^{2+3} \cdot 3 = -3 \]

- **Cofactor \( C_{31} \)**:
\[ M_{31} = \begin{vmatrix} -2 & -2 \\ 1 & -2 \end{vmatrix} = (-2)(-2) - (-2)(1) = 4 + 2 = 6 \]
\[ C_{31} = (-1)^{3+1} \cdot 6 = 6 \]

- **Cofactor \( C_{32} \)**:
\[ M_{32} = \begin{vmatrix} -1 & -2 \\ 2 & -2 \end{vmatrix} = (-1)(-2) - (-2)(2) = 2 + 4 = 6 \]
\[ C_{32} = (-1)^{3+2} \cdot 6 = -6 \]

- **Cofactor \( C_{33} \)**:
\[ M_{33} = \begin{vmatrix} -1 & -2 \\ 2 & 1 \end{vmatrix} = (-1)(1) - (-2)(2) = -1 + 4 = 3 \]
\[ C_{33} = (-1)^{3+3} \cdot 3 = 3 \]

Now, we can construct the cofactor matrix:
\[ \text{Cofactor Matrix} = \begin{pmatrix} -3 & -6 & -6 \\ 6 & 6 & -3 \\ 6 & -6 & 3 \end{pmatrix} \]

3. **Transpose the cofactor matrix** to get the adjoint of \( A \):
\[ \text{adj}(A) = \text{Cofactor Matrix}^T = \begin{pmatrix} -3 & 6 & 6 \\ -6 & 6 & -6 \\ -6 & -3 & 3 \end{pmatrix} \]

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