The inverse of the matrix `[[3, -2, -1], [-4, 1, -1], [2, 0, 1]]` is

Options

A. `[[1, 2, 3], [3, 3, 7], [-2, -4, -5]]`
B. `[[1, 2, 3], [2, 5, 7], [-2, -4, -5]]`
C. `[[1, -3, 5], [7, 4, 6], [4, 2, 7]]`
D. `[[1, 2, -4], [8, -4, -5], [3, 5, 2]]`

Answer

To find the inverse of the matrix

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

we will follow these steps:

### Step 1: Calculate the Determinant of A

The determinant of a 3x3 matrix

\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

is given by the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg). \]

For our matrix \( A \):

- \( a = 3, b = -2, c = -1 \)
- \( d = -4, e = 1, f = -1 \)
- \( g = 2, h = 0, i = 1 \)

Calculating the determinant:

\[ \text{det}(A) = 3(1 \cdot 1 - (-1) \cdot 0) - (-2)(-4 \cdot 1 - (-1) \cdot 2) + (-1)(-4 \cdot 0 - 1 \cdot 2) \]

\[ = 3(1 - 0) - (-2)(-4 + 2) - 1(0 - 2) \]

\[ = 3 - (-2)(-2) + 2 \]

\[ = 3 - 4 + 2 = 1. \]

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

Since \(\text{det}(A) = 1\) (which is not equal to 0), the inverse of matrix \( A \) exists.

### Step 3: Calculate the Adjoint of A

The adjoint of a matrix is the transpose of its cofactor matrix. We will calculate the cofactor matrix first.

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = \text{det}\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} = (1)(1) - (-1)(0) = 1. \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = -\text{det}\begin{bmatrix} -4 & -1 \\ 2 & 1 \end{bmatrix} = -((-4)(1) - (-1)(2)) = -(-4 + 2) = 2. \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = \text{det}\begin{bmatrix} -4 & 1 \\ 2 & 0 \end{bmatrix} = (-4)(0) - (1)(2) = -2. \]

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

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = \text{det}\begin{bmatrix} 3 & -1 \\ 2 & 1 \end{bmatrix} = (3)(1) - (-1)(2) = 3 + 2 = 5. \]

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

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

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = -\text{det}\begin{bmatrix} 3 & -1 \\ -4 & -1 \end{bmatrix} = -((3)(-1) - (-1)(-4)) = -(-3 - 4) = 7. \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = \text{det}\begin{bmatrix} 3 & -2 \\ -4 & 1 \end{bmatrix} = (3)(1) - (-2)(-4) = 3 - 8 = -5. \]

Now, the cofactor matrix \( C \) is:

\[ C = \begin{bmatrix} 1 & 2 & -2 \\ 2 & 5 & -4 \\ 3 & 7 & -5 \end{bmatrix}. \]

Taking the transpose to get the adjoint:

\[ \text{adj}(A) = C^T = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix}. \]

### Step 4: Calculate the Inverse of A

Using the formula for the inverse of a matrix:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A). \]

Since \(\text{det}(A) = 1\):

\[ A^{-1} = 1 \cdot \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix}. \]

### Final Answer

The inverse of the matrix \( A \) is

\[ A^{-1} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix}. \]