If `A=[(3,4,1),(1,0,-2),(-2,-1,2)] then A^(-1)`=?

Options

A. `[(2,9,-8),(-2,8,7),(-1,5,-4)]`
B. `[(2,9,-8),(2,8,7),(-1,-5,4)]`
C. `[(-2,-9,-8),(2,8,7),(-1,5,-4)]`
D. none of these

Explanation

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

### Step 1: Calculate the Determinant of \( A \)

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 \( A \):

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

Calculating each term:

1. \( 3(0 - 2) = 3 \cdot -2 = -6 \)
2. \( -4(2 - 4) = -4 \cdot -2 = 8 \)
3. \( 1(-1 - 0) = -1 \)

Now, summing these results:

\[ \text{det}(A) = -6 + 8 - 1 = 1 \]

### Step 2: Calculate the Adjoint of \( A \)

The adjoint of a matrix is the transpose of the cofactor matrix. We will calculate the cofactors for each element of \( A \).

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = (-1)^{1+1} \cdot \text{det} \begin{pmatrix} 0 & -2 \\ -1 & 2 \end{pmatrix} = 1(0 \cdot 2 - (-2)(-1)) = 0 - 2 = -2 \]

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

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

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = (-1)^{2+1} \cdot \text{det} \begin{pmatrix} 4 & 1 \\ -1 & 2 \end{pmatrix} = -1(4 \cdot 2 - 1 \cdot -1) = -1(8 + 1) = -9 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = (-1)^{2+2} \cdot \text{det} \begin{pmatrix} 3 & 1 \\ -2 & 2 \end{pmatrix} = 1(3 \cdot 2 - 1 \cdot -2) = 6 + 2 = 8 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = (-1)^{2+3} \cdot \text{det} \begin{pmatrix} 3 & 4 \\ -2 & -1 \end{pmatrix} = -1(3 \cdot -1 - 4 \cdot -2) = -1(-3 + 8) = -5 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = (-1)^{3+1} \cdot \text{det} \begin{pmatrix} 4 & 1 \\ 0 & -2 \end{pmatrix} = 1(4 \cdot -2 - 1 \cdot 0) = -8 \]

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = (-1)^{3+2} \cdot \text{det} \begin{pmatrix} 3 & 1 \\ 1 & -2 \end{pmatrix} = -1(3 \cdot -2 - 1 \cdot 1) = -1(-6 - 1) = 7 \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = (-1)^{3+3} \cdot \text{det} \begin{pmatrix} 3 & 4 \\ 1 & 0 \end{pmatrix} = 1(3 \cdot 0 - 4 \cdot 1) = -4 \]

The cofactor matrix is:

\[ \text{Cofactor}(A) = \begin{pmatrix} -2 & 2 & -1 \\ -9 & 8 & -5 \\ -8 & 7 & -4 \end{pmatrix} \]

Now, taking the transpose to get the adjoint:

\[ \text{adj}(A) = \begin{pmatrix} -2 & -9 & -8 \\ 2 & 8 & 7 \\ -1 & -5 & -4 \end{pmatrix} \]

### Step 3: Calculate the Inverse of \( A \)

Using the formula for the inverse:

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

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

\[ A^{-1} = 1 \cdot \begin{pmatrix} -2 & -9 & -8 \\ 2 & 8 & 7 \\ -1 & -5 & -4 \end{pmatrix} \]

Thus, the inverse of \( A \) is:

\[ A^{-1} = \begin{pmatrix} -2 & -9 & -8 \\ 2 & 8 & 7 \\ -1 & -5 & -4 \end{pmatrix} \]