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

Options

A. `(-1)/(5)[[-2, 0, 3], [-1, 1, 0], [2, 1,-2]]`
B. `(1)/(5)[[-2, 0, 3], [-1, 1, 0], [2, 1,-2]]`
C. `[[2, 0, -3], [1, -1, 0], [-2, -1, 2]]`
D. `[[-2, 0, 3], [-1, 1, 0], [2, 1, -2]]`

Answer

To find the inverse of the matrix \( A = \begin{bmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{bmatrix} \), 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 \( A \):

\[ \text{det}(A) = 2 \begin{vmatrix} 2 & 3 \\ -2 & 2 \end{vmatrix} - (-3) \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 2 \\ 3 & -2 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 3 \\ -2 & 2 \end{vmatrix} = (2)(2) - (3)(-2) = 4 + 6 = 10 \)
2. \( \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = (2)(2) - (3)(3) = 4 - 9 = -5 \)
3. \( \begin{vmatrix} 2 & 2 \\ 3 & -2 \end{vmatrix} = (2)(-2) - (2)(3) = -4 - 6 = -10 \)

Substituting back into the determinant formula:

\[ \text{det}(A) = 2(10) + 3(5) + 3(-10) = 20 + 15 - 30 = 5 \]

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

Since \( \text{det}(A) = 5 \neq 0 \), the inverse of matrix \( A \) exists.

### Step 3: Calculate the Adjoint of Matrix A

The adjoint of a matrix is the transpose of its cofactor matrix. We will calculate the cofactor matrix \( C \) first.

#### Calculate Cofactors:

1. \( C_{11} = \begin{vmatrix} 2 & 3 \\ -2 & 2 \end{vmatrix} = 10 \)
2. \( C_{12} = -\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = 5 \)
3. \( C_{13} = \begin{vmatrix} 2 & 2 \\ 3 & -2 \end{vmatrix} = -10 \)
4. \( C_{21} = -\begin{vmatrix} -3 & 3 \\ -2 & 2 \end{vmatrix} = 0 \)
5. \( C_{22} = \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = -5 \)
6. \( C_{23} = -\begin{vmatrix} 2 & -3 \\ 3 & -2 \end{vmatrix} = -5 \)
7. \( C_{31} = \begin{vmatrix} -3 & 3 \\ 2 & 3 \end{vmatrix} = -15 \)
8. \( C_{32} = -\begin{vmatrix} 2 & -3 \\ 2 & 3 \end{vmatrix} = 0 \)
9. \( C_{33} = \begin{vmatrix} 2 & -3 \\ 2 & 2 \end{vmatrix} = 10 \)

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

\[ C = \begin{bmatrix} 10 & 5 & -10 \\ 0 & -5 & -5 \\ -15 & 0 & 10 \end{bmatrix} \]

Now, we take the transpose of \( C \) to get the adjoint \( \text{adj}(A) \):

\[ \text{adj}(A) = C^T = \begin{bmatrix} 10 & 0 & -15 \\ 5 & -5 & 0 \\ -10 & -5 & 10 \end{bmatrix} \]

### Step 4: 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:

\[ A^{-1} = \frac{1}{5} \begin{bmatrix} 10 & 0 & -15 \\ 5 & -5 & 0 \\ -10 & -5 & 10 \end{bmatrix} \]

This simplifies to:

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

### Final Answer

The inverse of the matrix \( A \) is:

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