The inverse of the matrix `[[0, 1, 2], [1, 2, 3], [3, 1, 1]]` is

Options

A. `(-1)/(2)[[-1, 1, -1], [8, -6, 2], [-5, 3, -1]]`
B. `(1)/(2)[[-1, 1, -1], [8, -6, 1], [-5, 3, -1]]`
C. `[[-1, 1, -1], [4, -3, 1], [-5, 3, -1]]`
D. `[[1, -1, 1], [-4, 3, -1], [4, -3, 1]]`

Answer

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

### Step 1: Calculate the Determinant of Matrix A

The determinant of a 3x3 matrix \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \) is given by the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:
- \( a = 0, b = 1, c = 2 \)
- \( d = 1, e = 2, f = 3 \)
- \( g = 3, h = 1, i = 1 \)

Calculating the determinant:
\[ \text{det}(A) = 0(2 \cdot 1 - 3 \cdot 1) - 1(1 \cdot 1 - 3 \cdot 2) + 2(1 \cdot 1 - 2 \cdot 3) \]
\[ = 0 - 1(1 - 6) + 2(1 - 6) \]
\[ = 0 - 1(-5) + 2(-5) \]
\[ = 5 - 10 = -5 \]

### Step 2: Check if the Determinant is Non-Zero

Since \( \text{det}(A) = -5 \) (which is not equal to 0), the inverse of matrix \( A \) exists.

### Step 3: Calculate the Cofactor Matrix

To find the cofactor matrix, we need to calculate the cofactor for each element of the matrix.

1. **Cofactor \( C_{1,1} \)**:
\[ C_{1,1} = \text{det}\begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} = (2)(1) - (3)(1) = 2 - 3 = -1 \]

2. **Cofactor \( C_{1,2} \)**:
\[ C_{1,2} = -\text{det}\begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix} = -((1)(1) - (3)(3)) = -(1 - 9) = 8 \]

3. **Cofactor \( C_{1,3} \)**:
\[ C_{1,3} = \text{det}\begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} = (1)(1) - (2)(3) = 1 - 6 = -5 \]

4. **Cofactor \( C_{2,1} \)**:
\[ C_{2,1} = -\text{det}\begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} = -((1)(1) - (2)(1)) = -(1 - 2) = 1 \]

5. **Cofactor \( C_{2,2} \)**:
\[ C_{2,2} = \text{det}\begin{bmatrix} 0 & 2 \\ 3 & 1 \end{bmatrix} = (0)(1) - (2)(3) = 0 - 6 = -6 \]

6. **Cofactor \( C_{2,3} \)**:
\[ C_{2,3} = -\text{det}\begin{bmatrix} 0 & 1 \\ 3 & 1 \end{bmatrix} = -((0)(1) - (1)(3)) = -(-3) = 3 \]

7. **Cofactor \( C_{3,1} \)**:
\[ C_{3,1} = \text{det}\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = (1)(3) - (2)(2) = 3 - 4 = -1 \]

8. **Cofactor \( C_{3,2} \)**:
\[ C_{3,2} = -\text{det}\begin{bmatrix} 0 & 2 \\ 1 & 3 \end{bmatrix} = -((0)(3) - (2)(1)) = -(-2) = 2 \]

9. **Cofactor \( C_{3,3} \)**:
\[ C_{3,3} = \text{det}\begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix} = (0)(2) - (1)(1) = 0 - 1 = -1 \]

The cofactor matrix \( C \) is:
\[ C = \begin{bmatrix} -1 & 8 & -5 \\ 1 & -6 & 3 \\ -1 & 2 & -1 \end{bmatrix} \]

### Step 4: Transpose the Cofactor Matrix to Get the Adjoint Matrix

The adjoint matrix \( \text{adj}(A) \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = C^T = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \]

### Step 5: Calculate the Inverse of Matrix A

Using the formula for the inverse:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values:
\[ A^{-1} = \frac{1}{-5} \cdot \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \]
\[ = \begin{bmatrix} \frac{1}{5} & -\frac{1}{5} & \frac{1}{5} \\ -\frac{8}{5} & \frac{6}{5} & -\frac{2}{5} \\ 1 & -\frac{3}{5} & \frac{1}{5} \end{bmatrix} \]

### Final Answer
Thus, the inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{bmatrix} \frac{1}{5} & -\frac{1}{5} & \frac{1}{5} \\ -\frac{8}{5} & \frac{6}{5} & -\frac{2}{5} \\ 1 & -\frac{3}{5} & \frac{1}{5} \end{bmatrix} \]