If `A=[[3, 5, -1], [2, 0, 4], [1,-3, 0]]`, then `[[12, 4, -6], [3, 1, 14], [20, -14, -10]]` is

Options

A. adj(A')
B. `-(adj(A))`
C. `adjA`
D. `-A^(-1)`

Answer

To solve the problem, we need to find the adjoint of the matrix \( A \) given by:

\[ A = \begin{bmatrix} 3 & 5 & -1 \\ 2 & 0 & 4 \\ 1 & -3 & 0 \end{bmatrix} \]

### Step 1: Calculate the Cofactor Matrix

The cofactor matrix is obtained by calculating the determinant of the 2x2 submatrices formed by removing one row and one column for each element of the matrix, while also applying the appropriate sign based on the position of the element.

1. **Cofactor \( C_{11} \)**:
- Remove the first row and first column:
\[ \begin{vmatrix} 0 & 4 \\ -3 & 0 \end{vmatrix} = (0)(0) - (4)(-3) = 12 \quad \text{(positive)} \]
Thus, \( C_{11} = 12 \).

2. **Cofactor \( C_{12} \)**:
- Remove the first row and second column:
\[ \begin{vmatrix} 2 & 4 \\ 1 & 0 \end{vmatrix} = (2)(0) - (4)(1) = -4 \quad \text{(negative)} \]
Thus, \( C_{12} = 4 \).

3. **Cofactor \( C_{13} \)**:
- Remove the first row and third column:
\[ \begin{vmatrix} 2 & 0 \\ 1 & -3 \end{vmatrix} = (2)(-3) - (0)(1) = -6 \quad \text{(positive)} \]
Thus, \( C_{13} = -6 \).

4. **Cofactor \( C_{21} \)**:
- Remove the second row and first column:
\[ \begin{vmatrix} 5 & -1 \\ -3 & 0 \end{vmatrix} = (5)(0) - (-1)(-3) = -3 \quad \text{(negative)} \]
Thus, \( C_{21} = 3 \).

5. **Cofactor \( C_{22} \)**:
- Remove the second row and second column:
\[ \begin{vmatrix} 3 & -1 \\ 1 & 0 \end{vmatrix} = (3)(0) - (-1)(1) = 1 \quad \text{(positive)} \]
Thus, \( C_{22} = 1 \).

6. **Cofactor \( C_{23} \)**:
- Remove the second row and third column:
\[ \begin{vmatrix} 3 & 5 \\ 1 & -3 \end{vmatrix} = (3)(-3) - (5)(1) = -9 - 5 = -14 \quad \text{(negative)} \]
Thus, \( C_{23} = 14 \).

7. **Cofactor \( C_{31} \)**:
- Remove the third row and first column:
\[ \begin{vmatrix} 5 & -1 \\ 0 & 4 \end{vmatrix} = (5)(4) - (-1)(0) = 20 \quad \text{(positive)} \]
Thus, \( C_{31} = 20 \).

8. **Cofactor \( C_{32} \)**:
- Remove the third row and second column:
\[ \begin{vmatrix} 3 & -1 \\ 2 & 4 \end{vmatrix} = (3)(4) - (-1)(2) = 12 + 2 = 14 \quad \text{(negative)} \]
Thus, \( C_{32} = -14 \).

9. **Cofactor \( C_{33} \)**:
- Remove the third row and third column:
\[ \begin{vmatrix} 3 & 5 \\ 2 & 0 \end{vmatrix} = (3)(0) - (5)(2) = -10 \quad \text{(positive)} \]
Thus, \( C_{33} = -10 \).

### Step 2: Form the Cofactor Matrix

Now we can form the cofactor matrix \( C \):

\[ C = \begin{bmatrix} 12 & 4 & -6 \\ 3 & 1 & 14 \\ 20 & -14 & -10 \end{bmatrix} \]

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

The adjoint of matrix \( A \) is the transpose of the cofactor matrix:

\[ \text{Adjoint } A = C^T = \begin{bmatrix} 12 & 3 & 20 \\ 4 & 1 & -14 \\ -6 & 14 & -10 \end{bmatrix} \]

### Step 4: Compare with Given Matrix

The given matrix is:

\[ \begin{bmatrix} 12 & 4 & -6 \\ 3 & 1 & 14 \\ 20 & -14 & -10 \end{bmatrix} \]

### Conclusion

The given matrix is actually the adjoint of \( A \) transposed, which means:

\[ \text{Given Matrix} = \text{Adjoint } A^T \]

Thus, the answer is the first option: **Adjoint \( A^T \)**.