Adjoint of the matrix `N=[[-4, -3, -3], [1, 0, 1], [4, 4, 3]]` is

Options

A. `-N`
B. `N`
C. `2N`
D. `-2N`

Answer

To find the adjoint of the matrix \( N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix} \), we will follow these steps:

### Step 1: Find the Cofactor Matrix
The cofactor matrix \( C \) is obtained by calculating the cofactors of each element in the matrix \( N \).

#### Cofactor Calculation:
1. **Cofactor \( C_{11} \)**:
- Minor of \( C_{11} \) (determinant of the matrix obtained by deleting the 1st row and 1st column):
\[ \text{Minor} = \begin{vmatrix} 0 & 1 \\ 4 & 3 \end{vmatrix} = (0 \cdot 3 - 1 \cdot 4) = -4 \]
- Therefore, \( C_{11} = (-1)^{1+1} \cdot (-4) = -4 \).

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

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

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

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

6. **Cofactor \( C_{23} \)**:
- Minor of \( C_{23} \):
\[ \text{Minor} = \begin{vmatrix} -4 & -3 \\ 4 & 4 \end{vmatrix} = (-4 \cdot 4 - (-3) \cdot 4) = -16 + 12 = -4 \]
- Therefore, \( C_{23} = (-1)^{2+3} \cdot (-4) = 4 \).

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

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

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

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

### Step 3: Transpose the Cofactor Matrix
To find the adjoint, we take the transpose of the cofactor matrix:
\[ \text{adj}(N) = C^T = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix} \]

### Step 4: Compare with Original Matrix
Notice that:
\[ \text{adj}(N) = N \]

### Final Answer
Thus, the adjoint of the matrix \( N \) is:
\[ \text{adj}(N) = N \]