Inverse of the matrix `[(3,-2,-1),(-4,1,-1),(2,0,1)]` is

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of the Matrix
The determinant of a 3x3 matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) 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:
\[ \text{det}(A) = 3(1 \cdot 1 - (-1) \cdot 0) - (-2)(-4 \cdot 1 - (-1) \cdot 2) + (-1)(-4 \cdot 0 - 1 \cdot 2) \]
\[ = 3(1) - (-2)(-4 + 2) - 1(-2) \]
\[ = 3 - (-2)(-2) + 2 \]
\[ = 3 - 4 + 2 = 1 \]

### Step 2: Calculate the Cofactor Matrix
To find the cofactor matrix, we need to calculate the cofactor for each element of the matrix \( A \).

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

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

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

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

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

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

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

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

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

The cofactor matrix is:
\[ C = \begin{pmatrix} 1 & 2 & -2 \\ 2 & 5 & -4 \\ 3 & 7 & -5 \end{pmatrix} \]

### Step 3: Transpose the Cofactor Matrix
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = C^T = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix} \]

### Step 4: Calculate the Inverse of the Matrix
The inverse of matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Since \( \text{det}(A) = 1 \):
\[ A^{-1} = 1 \cdot \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix} \]

### Final Answer:
The inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix} \]