If `A=[(0,1,1),(1,2,0),(4,-1,3)]` then Adj. A=……….

Answer

To find the adjoint of the matrix \( A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 2 & 0 \\ 4 & -1 & 3 \end{pmatrix} \), we will follow these steps:

### Step 1: Find the Cofactor Matrix

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

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

where \( M_{ij} \) is the minor of the element \( a_{ij} \), which is the determinant of the matrix obtained by deleting the \( i \)-th row and \( j \)-th column from \( A \).

#### Calculation of Cofactors:

1. **Cofactor \( C_{11} \)**:
- Minor \( M_{11} = \det\begin{pmatrix} 2 & 0 \\ -1 & 3 \end{pmatrix} = (2)(3) - (0)(-1) = 6 \)
- \( C_{11} = (-1)^{1+1} M_{11} = 1 \cdot 6 = 6 \)

2. **Cofactor \( C_{12} \)**:
- Minor \( M_{12} = \det\begin{pmatrix} 1 & 0 \\ 4 & 3 \end{pmatrix} = (1)(3) - (0)(4) = 3 \)
- \( C_{12} = (-1)^{1+2} M_{12} = -1 \cdot 3 = -3 \)

3. **Cofactor \( C_{13} \)**:
- Minor \( M_{13} = \det\begin{pmatrix} 1 & 2 \\ 4 & -1 \end{pmatrix} = (1)(-1) - (2)(4) = -1 - 8 = -9 \)
- \( C_{13} = (-1)^{1+3} M_{13} = 1 \cdot (-9) = -9 \)

4. **Cofactor \( C_{21} \)**:
- Minor \( M_{21} = \det\begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix} = (1)(3) - (1)(-1) = 3 + 1 = 4 \)
- \( C_{21} = (-1)^{2+1} M_{21} = -1 \cdot 4 = -4 \)

5. **Cofactor \( C_{22} \)**:
- Minor \( M_{22} = \det\begin{pmatrix} 0 & 1 \\ 4 & 3 \end{pmatrix} = (0)(3) - (1)(4) = -4 \)
- \( C_{22} = (-1)^{2+2} M_{22} = 1 \cdot (-4) = -4 \)

6. **Cofactor \( C_{23} \)**:
- Minor \( M_{23} = \det\begin{pmatrix} 0 & 1 \\ 4 & -1 \end{pmatrix} = (0)(-1) - (1)(4) = -4 \)
- \( C_{23} = (-1)^{2+3} M_{23} = -1 \cdot (-4) = 4 \)

7. **Cofactor \( C_{31} \)**:
- Minor \( M_{31} = \det\begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix} = (1)(0) - (1)(2) = -2 \)
- \( C_{31} = (-1)^{3+1} M_{31} = 1 \cdot (-2) = -2 \)

8. **Cofactor \( C_{32} \)**:
- Minor \( M_{32} = \det\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = (0)(0) - (1)(1) = -1 \)
- \( C_{32} = (-1)^{3+2} M_{32} = -1 \cdot (-1) = 1 \)

9. **Cofactor \( C_{33} \)**:
- Minor \( M_{33} = \det\begin{pmatrix} 0 & 1 \\ 1 & 2 \end{pmatrix} = (0)(2) - (1)(1) = -1 \)
- \( C_{33} = (-1)^{3+3} M_{33} = 1 \cdot (-1) = -1 \)

#### Cofactor Matrix:

Now, we can write the cofactor matrix \( C \):

\[ C = \begin{pmatrix} 6 & -3 & -9 \\ -4 & -4 & 4 \\ -2 & 1 & -1 \end{pmatrix} \]

### Step 2: Transpose the Cofactor Matrix

To find the adjoint, we need to transpose the cofactor matrix:

\[ \text{Adj}(A) = C^T = \begin{pmatrix} 6 & -4 & -2 \\ -3 & -4 & 1 \\ -9 & 4 & -1 \end{pmatrix} \]

### Final Answer:

Thus, the adjoint of matrix \( A \) is:

\[ \text{Adj}(A) = \begin{pmatrix} 6 & -4 & -2 \\ -3 & -4 & 1 \\ -9 & 4 & -1 \end{pmatrix} \]

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