The adjoint matrix of `[(3,-3,4),(2,-3,4),(0,-1,1)]` is

Options

A. `[(4,8,3),(2,1,6),(0,2,1)]`
B. `[(1,-1,0),(-2,3,-4),(-2,3,-3)]`
C. `[(11,9,3),(1,2,8),(6,9,1)]`
D. `[(1,-2,1),(-1,3,3),(-2,3,-3)]`

Answer

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

### Step 1: Find the Matrix of Minors

The first step in finding the adjoint is to calculate the matrix of minors. The minor of an element in a matrix is the determinant of the submatrix that remains after removing the row and column of that element.

1. **Minor of \( a_{11} = 3 \)**:
\[ M_{11} = \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = (-3)(1) - (4)(-1) = -3 + 4 = 1 \]

2. **Minor of \( a_{12} = -3 \)**:
\[ M_{12} = \begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} = (2)(1) - (4)(0) = 2 - 0 = 2 \]

3. **Minor of \( a_{13} = 4 \)**:
\[ M_{13} = \begin{vmatrix} 2 & -3 \\ 0 & -1 \end{vmatrix} = (2)(-1) - (-3)(0) = -2 - 0 = -2 \]

4. **Minor of \( a_{21} = 2 \)**:
\[ M_{21} = \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = (-3)(1) - (4)(-1) = -3 + 4 = 1 \]

5. **Minor of \( a_{22} = -3 \)**:
\[ M_{22} = \begin{vmatrix} 3 & 4 \\ 0 & 1 \end{vmatrix} = (3)(1) - (4)(0) = 3 - 0 = 3 \]

6. **Minor of \( a_{23} = 4 \)**:
\[ M_{23} = \begin{vmatrix} 3 & -3 \\ 0 & -1 \end{vmatrix} = (3)(-1) - (-3)(0) = -3 - 0 = -3 \]

7. **Minor of \( a_{31} = 0 \)**:
\[ M_{31} = \begin{vmatrix} -3 & 4 \\ -3 & 4 \end{vmatrix} = (-3)(4) - (4)(-3) = -12 + 12 = 0 \]

8. **Minor of \( a_{32} = -1 \)**:
\[ M_{32} = \begin{vmatrix} 3 & 4 \\ 2 & 4 \end{vmatrix} = (3)(4) - (4)(2) = 12 - 8 = 4 \]

9. **Minor of \( a_{33} = 1 \)**:
\[ M_{33} = \begin{vmatrix} 3 & -3 \\ 2 & -3 \end{vmatrix} = (3)(-3) - (-3)(2) = -9 + 6 = -3 \]

Thus, the matrix of minors is:
\[ M = \begin{pmatrix} 1 & 2 & -2 \\ 1 & 3 & -3 \\ 0 & 4 & -3 \end{pmatrix} \]

### Step 2: Find the Matrix of Cofactors

Next, we apply the checkerboard pattern of signs to the matrix of minors to get the matrix of cofactors.

\[ C = \begin{pmatrix} +1 & -2 & +2 \\ -1 & +3 & -3 \\ +0 & -4 & +3 \end{pmatrix} = \begin{pmatrix} 1 & -2 & 2 \\ -1 & 3 & -3 \\ 0 & -4 & 3 \end{pmatrix} \]

### Step 3: Find the Adjoint Matrix

The adjoint of the matrix \( A \) is the transpose of the matrix of cofactors.

\[ \text{adj}(A) = C^T = \begin{pmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ 2 & -3 & 3 \end{pmatrix} \]

### Final Answer

Thus, the adjoint matrix of \( A \) is:
\[ \text{adj}(A) = \begin{pmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ 2 & -3 & 3 \end{pmatrix} \]

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