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

Options

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

Answer

To find the inverse of the matrix \( A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix} \), 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 calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \( A \):
- \( a = 1, b = 0, c = 1 \)
- \( d = 0, e = 2, f = 3 \)
- \( g = 1, h = 2, i = 1 \)

Calculating the determinant:
\[ \text{det}(A) = 1(2 \cdot 1 - 3 \cdot 2) - 0(0 \cdot 1 - 3 \cdot 1) + 1(0 \cdot 2 - 2 \cdot 1) \]
\[ = 1(2 - 6) + 0 + 1(0 - 2) \]
\[ = 1(-4) + 0 - 2 = -4 - 2 = -6 \]

### Step 2: Check if the Determinant is Non-Zero
Since \( \text{det}(A) = -6 \neq 0 \), the inverse of matrix \( A \) exists.

### Step 3: Calculate the Cofactor Matrix
The cofactor matrix \( C \) is calculated by finding the determinant of the 2x2 submatrices formed by removing the row and column of each element, and applying the checkerboard pattern of signs.

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = \text{det}\begin{bmatrix} 2 & 3 \\ 2 & 1 \end{bmatrix} = (2 \cdot 1 - 3 \cdot 2) = 2 - 6 = -4 \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = -\text{det}\begin{bmatrix} 0 & 3 \\ 1 & 1 \end{bmatrix} = -(0 \cdot 1 - 3 \cdot 1) = 3 \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = \text{det}\begin{bmatrix} 0 & 2 \\ 1 & 2 \end{bmatrix} = (0 \cdot 2 - 2 \cdot 1) = -2 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = -\text{det}\begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} = -(0 \cdot 1 - 1 \cdot 2) = 2 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = \text{det}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = (1 \cdot 1 - 1 \cdot 1) = 0 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = -\text{det}\begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} = -(1 \cdot 2 - 0 \cdot 1) = -2 \]

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

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = -\text{det}\begin{bmatrix} 1 & 1 \\ 0 & 3 \end{bmatrix} = -(1 \cdot 3 - 1 \cdot 0) = -3 \]

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

Thus, the cofactor matrix \( C \) is:
\[ C = \begin{bmatrix} -4 & 3 & -2 \\ 2 & 0 & -2 \\ -2 & -3 & 2 \end{bmatrix} \]

### Step 4: Calculate the Adjoint of Matrix A
The adjoint of matrix \( A \) is the transpose of the cofactor matrix \( C \):
\[ \text{adj}(A) = C^T = \begin{bmatrix} -4 & 2 & -2 \\ 3 & 0 & -3 \\ -2 & -2 & 2 \end{bmatrix} \]

### 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 we found:
\[ A^{-1} = \frac{1}{-6} \cdot \begin{bmatrix} -4 & 2 & -2 \\ 3 & 0 & -3 \\ -2 & -2 & 2 \end{bmatrix} \]
\[ = \begin{bmatrix} \frac{4}{6} & -\frac{2}{6} & \frac{2}{6} \\ -\frac{3}{6} & 0 & \frac{3}{6} \\ \frac{2}{6} & \frac{2}{6} & -\frac{2}{6} \end{bmatrix} \]
\[ = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} \end{bmatrix} \]

### Final Answer
The inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} \end{bmatrix} \]