If `A = [(-4,-3,-3),(1,0,1),(4,4,3)]`, find adj (A)

Answer

To find the adjoint of the matrix \( A = \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{pmatrix} \), we need to calculate the cofactors of each element of the matrix and then take the transpose of the cofactor matrix.

### Step 1: Calculate the Cofactors

The cofactor \( C_{ij} \) of an element \( a_{ij} \) 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 \( A \).

#### Calculate \( C_{11} \)
To find \( C_{11} \):
- Remove the first row and first column:
\[ M_{11} = \begin{pmatrix} 0 & 1 \\ 4 & 3 \end{pmatrix} \]
- Calculate the determinant:
\[ \text{det}(M_{11}) = (0)(3) - (1)(4) = -4 \]
- Therefore,
\[ C_{11} = (-1)^{1+1} \cdot (-4) = -4 \]

#### Calculate \( C_{12} \)
To find \( C_{12} \):
- Remove the first row and second column:
\[ M_{12} = \begin{pmatrix} 1 & 1 \\ 4 & 3 \end{pmatrix} \]
- Calculate the determinant:
\[ \text{det}(M_{12}) = (1)(3) - (1)(4) = 3 - 4 = -1 \]
- Therefore,
\[ C_{12} = (-1)^{1+2} \cdot (-1) = 1 \]

#### Calculate \( C_{13} \)
To find \( C_{13} \):
- Remove the first row and third column:
\[ M_{13} = \begin{pmatrix} 1 & 0 \\ 4 & 4 \end{pmatrix} \]
- Calculate the determinant:
\[ \text{det}(M_{13}) = (1)(4) - (0)(4) = 4 \]
- Therefore,
\[ C_{13} = (-1)^{1+3} \cdot 4 = 4 \]

#### Calculate \( C_{21} \)
To find \( C_{21} \):
- Remove the second row and first column:
\[ M_{21} = \begin{pmatrix} -3 & -3 \\ 4 & 3 \end{pmatrix} \]
- Calculate the determinant:
\[ \text{det}(M_{21}) = (-3)(3) - (-3)(4) = -9 + 12 = 3 \]
- Therefore,
\[ C_{21} = (-1)^{2+1} \cdot 3 = -3 \]

#### Calculate \( C_{22} \)
To find \( C_{22} \):
- Remove the second row and second column:
\[ M_{22} = \begin{pmatrix} -4 & -3 \\ 4 & 3 \end{pmatrix} \]
- Calculate the determinant:
\[ \text{det}(M_{22}) = (-4)(3) - (-3)(4) = -12 + 12 = 0 \]
- Therefore,
\[ C_{22} = (-1)^{2+2} \cdot 0 = 0 \]

#### Calculate \( C_{23} \)
To find \( C_{23} \):
- Remove the second row and third column:
\[ M_{23} = \begin{pmatrix} -4 & -3 \\ 4 & 4 \end{pmatrix} \]
- Calculate the determinant:
\[ \text{det}(M_{23}) = (-4)(4) - (-3)(4) = -16 + 12 = -4 \]
- Therefore,
\[ C_{23} = (-1)^{2+3} \cdot (-4) = 4 \]

#### Calculate \( C_{31} \)
To find \( C_{31} \):
- Remove the third row and first column:
\[ M_{31} = \begin{pmatrix} -3 & -3 \\ 0 & 1 \end{pmatrix} \]
- Calculate the determinant:
\[ \text{det}(M_{31}) = (-3)(1) - (-3)(0) = -3 \]
- Therefore,
\[ C_{31} = (-1)^{3+1} \cdot (-3) = -3 \]

#### Calculate \( C_{32} \)
To find \( C_{32} \):
- Remove the third row and second column:
\[ M_{32} = \begin{pmatrix} -4 & -3 \\ 1 & 1 \end{pmatrix} \]
- Calculate the determinant:
\[ \text{det}(M_{32}) = (-4)(1) - (-3)(1) = -4 + 3 = -1 \]
- Therefore,
\[ C_{32} = (-1)^{3+2} \cdot (-1) = 1 \]

#### Calculate \( C_{33} \)
To find \( C_{33} \):
- Remove the third row and third column:
\[ M_{33} = \begin{pmatrix} -4 & -3 \\ 1 & 0 \end{pmatrix} \]
- Calculate the determinant:
\[ \text{det}(M_{33}) = (-4)(0) - (-3)(1) = 0 + 3 = 3 \]
- Therefore,
\[ C_{33} = (-1)^{3+3} \cdot 3 = 3 \]

### Step 2: Form the Cofactor Matrix
Now we can form the cofactor matrix \( C \):
\[ C = \begin{pmatrix} -4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & 3 \end{pmatrix} \]

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

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