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

Answer

To solve the problem, we need to show that the adjoint of matrix \( A \) is equal to \( A \) itself, where

\[ A = \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{pmatrix} \]

### Step 1: Calculate the Cofactor Matrix

The cofactor \( C_{ij} \) of an element \( a_{ij} \) in matrix \( A \) is given by:

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

where \( M_{ij} \) is the determinant of the submatrix formed by deleting the \( i \)-th row and \( j \)-th column.

#### Finding Cofactors:

1. **Cofactor \( C_{11} \)**:
\[ M_{11} = \begin{vmatrix} 0 & 1 \\ 4 & 3 \end{vmatrix} = (0 \cdot 3) - (1 \cdot 4) = -4 \]
\[ C_{11} = (-1)^{1+1} \cdot (-4) = -4 \]

2. **Cofactor \( C_{12} \)**:
\[ M_{12} = \begin{vmatrix} 1 & 1 \\ 4 & 3 \end{vmatrix} = (1 \cdot 3) - (1 \cdot 4) = -1 \]
\[ C_{12} = (-1)^{1+2} \cdot (-1) = 1 \]

3. **Cofactor \( C_{13} \)**:
\[ M_{13} = \begin{vmatrix} 1 & 0 \\ 4 & 4 \end{vmatrix} = (1 \cdot 4) - (0 \cdot 4) = 4 \]
\[ C_{13} = (-1)^{1+3} \cdot 4 = 4 \]

4. **Cofactor \( C_{21} \)**:
\[ M_{21} = \begin{vmatrix} -3 & -3 \\ 4 & 3 \end{vmatrix} = (-3 \cdot 3) - (-3 \cdot 4) = -9 + 12 = 3 \]
\[ C_{21} = (-1)^{2+1} \cdot 3 = -3 \]

5. **Cofactor \( C_{22} \)**:
\[ M_{22} = \begin{vmatrix} -4 & -3 \\ 4 & 3 \end{vmatrix} = (-4 \cdot 3) - (-3 \cdot 4) = -12 + 12 = 0 \]
\[ C_{22} = (-1)^{2+2} \cdot 0 = 0 \]

6. **Cofactor \( C_{23} \)**:
\[ M_{23} = \begin{vmatrix} -4 & -3 \\ 1 & 4 \end{vmatrix} = (-4 \cdot 4) - (-3 \cdot 1) = -16 + 3 = -13 \]
\[ C_{23} = (-1)^{2+3} \cdot (-13) = 13 \]

7. **Cofactor \( C_{31} \)**:
\[ M_{31} = \begin{vmatrix} -3 & -3 \\ 0 & 1 \end{vmatrix} = (-3 \cdot 1) - (-3 \cdot 0) = -3 \]
\[ C_{31} = (-1)^{3+1} \cdot (-3) = -3 \]

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

9. **Cofactor \( C_{33} \)**:
\[ M_{33} = \begin{vmatrix} -4 & -3 \\ 1 & 0 \end{vmatrix} = (-4 \cdot 0) - (-3 \cdot 1) = 3 \]
\[ C_{33} = (-1)^{3+3} \cdot 3 = 3 \]

### Step 2: Form the Cofactor Matrix

The cofactor matrix \( C \) is:

\[ C = \begin{pmatrix} -4 & 1 & 4 \\ -3 & 0 & 13 \\ -3 & 1 & 3 \end{pmatrix} \]

### Step 3: Transpose the Cofactor Matrix to Get the Adjoint

The adjoint \( \text{adj}(A) \) is the transpose of the cofactor matrix:

\[ \text{adj}(A) = C^T = \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 13 & 3 \end{pmatrix} \]

### Step 4: Compare \( \text{adj}(A) \) with \( A \)

Now we need to check if \( \text{adj}(A) = A \):

\[ A = \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{pmatrix} \]

Since the entries of \( \text{adj}(A) \) do not match with \( A \), we find that \( \text{adj}(A) \neq A \).

### Conclusion

Thus, the statement \( \text{adj}(A) = A \) is not true based on our calculations.

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