If `A[(2,0,3),(1,-1,2),(3,-2,0)]`, then `(adj A)A=`

Options

A. `[(-11,0,0),(0,-11,0),(0,0,-11)]`
B. `[(11,0,0),(0,11,0),(0,0,11)]`
C. `[(17,0,0),(0,17,0),(0,0,17)]`
D. `[(-7,0,0),(0,-7,0),(0,0,-7)]`

Answer

To solve the problem, we need to find the product of the adjoint of matrix \( A \) and matrix \( A \) itself. Let's go through the steps one by one.

### Step 1: Define the Matrix \( A \)
Given the matrix \( A \):
\[ A = \begin{pmatrix} 2 & 0 & 3 \\ 1 & -1 & 2 \\ 3 & -2 & 0 \end{pmatrix} \]

### Step 2: Calculate the Cofactor Matrix \( C \)
To find the adjoint of \( A \), we first need to calculate the cofactor matrix \( C \). The cofactor \( C_{ij} \) is calculated as \( (-1)^{i+j} \) times the determinant of the minor matrix obtained by deleting the \( i^{th} \) row and \( j^{th} \) column.

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = \begin{vmatrix} -1 & 2 \\ -2 & 0 \end{vmatrix} = (-1)(0) - (2)(-2) = 0 + 4 = 4 \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = -\begin{vmatrix} 1 & 2 \\ 3 & 0 \end{vmatrix} = -[(1)(0) - (2)(3)] = -(-6) = 6 \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = \begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix} = (1)(-2) - (-1)(3) = -2 + 3 = 1 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = -\begin{vmatrix} 0 & 3 \\ -2 & 0 \end{vmatrix} = -[(0)(0) - (3)(-2)] = -6 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = \begin{vmatrix} 2 & 3 \\ 3 & 0 \end{vmatrix} = (2)(0) - (3)(3) = -9 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = -\begin{vmatrix} 2 & 0 \\ 3 & -2 \end{vmatrix} = -[(2)(-2) - (0)(3)] = 4 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = \begin{vmatrix} 0 & 3 \\ -1 & 2 \end{vmatrix} = (0)(2) - (3)(-1) = 3 \]

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

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = \begin{vmatrix} 2 & 0 \\ 1 & -1 \end{vmatrix} = (2)(-1) - (0)(1) = -2 \]

Now, the cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} 4 & 6 & 1 \\ -6 & -9 & 4 \\ 3 & -1 & -2 \end{pmatrix} \]

### Step 3: Calculate the Adjoint of \( A \)
The adjoint of \( A \) is the transpose of the cofactor matrix \( C \):
\[ \text{adj} A = C^T = \begin{pmatrix} 4 & -6 & 3 \\ 6 & -9 & -1 \\ 1 & 4 & -2 \end{pmatrix} \]

### Step 4: Calculate \( (\text{adj} A) A \)
Now we need to multiply \( \text{adj} A \) with \( A \):
\[ (\text{adj} A) A = \begin{pmatrix} 4 & -6 & 3 \\ 6 & -9 & -1 \\ 1 & 4 & -2 \end{pmatrix} \begin{pmatrix} 2 & 0 & 3 \\ 1 & -1 & 2 \\ 3 & -2 & 0 \end{pmatrix} \]

Calculating the product:
1. **First row, first column**:
\[ 4 \cdot 2 + (-6) \cdot 1 + 3 \cdot 3 = 8 - 6 + 9 = 11 \]

2. **First row, second column**:
\[ 4 \cdot 0 + (-6) \cdot (-1) + 3 \cdot (-2) = 0 + 6 - 6 = 0 \]

3. **First row, third column**:
\[ 4 \cdot 3 + (-6) \cdot 2 + 3 \cdot 0 = 12 - 12 + 0 = 0 \]

4. **Second row, first column**:
\[ 6 \cdot 2 + (-9) \cdot 1 + (-1) \cdot 3 = 12 - 9 - 3 = 0 \]

5. **Second row, second column**:
\[ 6 \cdot 0 + (-9) \cdot (-1) + (-1) \cdot (-2) = 0 + 9 + 2 = 11 \]

6. **Second row, third column**:
\[ 6 \cdot 3 + (-9) \cdot 2 + (-1) \cdot 0 = 18 - 18 + 0 = 0 \]

7. **Third row, first column**:
\[ 1 \cdot 2 + 4 \cdot 1 + (-2) \cdot 3 = 2 + 4 - 6 = 0 \]

8. **Third row, second column**:
\[ 1 \cdot 0 + 4 \cdot (-1) + (-2) \cdot (-2) = 0 - 4 + 4 = 0 \]

9. **Third row, third column**:
\[ 1 \cdot 3 + 4 \cdot 2 + (-2) \cdot 0 = 3 + 8 + 0 = 11 \]

Thus, the product \( (\text{adj} A) A \) is:
\[ (\text{adj} A) A = \begin{pmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{pmatrix} \]

### Final Answer
\[ (\text{adj} A) A = 11I \quad \text{where } I \text{ is the identity matrix.} \]