Using elementary transformations find the inverse of the following matrices `[{:( 2,-3,3),(2,2,3),(3,-2,2):}]`

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{pmatrix} \) using elementary transformations, we will augment the matrix \( A \) with the identity matrix \( I \) of the same order and then perform row operations to transform \( A \) into \( I \). The resulting augmented matrix will give us the inverse of \( A \).

### Step-by-Step Solution:

1. **Set up the augmented matrix**:
\[ \left( \begin{array}{ccc|ccc} 2 & -3 & 3 & 1 & 0 & 0 \\ 2 & 2 & 3 & 0 & 1 & 0 \\ 3 & -2 & 2 & 0 & 0 & 1 \end{array} \right) \]

2. **Perform the first row operation**: \( R_1 \leftarrow R_1 - R_3 \)
\[ \left( \begin{array}{ccc|ccc} -1 & -1 & 1 & 1 & 0 & -1 \\ 2 & 2 & 3 & 0 & 1 & 0 \\ 3 & -2 & 2 & 0 & 0 & 1 \end{array} \right) \]

3. **Multiply the first row by -1**: \( R_1 \leftarrow -1 \cdot R_1 \)
\[ \left( \begin{array}{ccc|ccc} 1 & 1 & -1 & -1 & 0 & 1 \\ 2 & 2 & 3 & 0 & 1 & 0 \\ 3 & -2 & 2 & 0 & 0 & 1 \end{array} \right) \]

4. **Perform the second row operation**: \( R_2 \leftarrow R_2 - 2R_1 \) and \( R_3 \leftarrow R_3 - 3R_1 \)
\[ \left( \begin{array}{ccc|ccc} 1 & 1 & -1 & -1 & 0 & 1 \\ 0 & 0 & 5 & 2 & 1 & -2 \\ 0 & -5 & 5 & 3 & 0 & -2 \end{array} \right) \]

5. **Perform the row operation**: \( R_3 \leftarrow R_3 + R_2 \)
\[ \left( \begin{array}{ccc|ccc} 1 & 1 & -1 & -1 & 0 & 1 \\ 0 & 0 & 5 & 2 & 1 & -2 \\ 0 & 0 & 10 & 5 & 1 & -4 \end{array} \right) \]

6. **Multiply the second row by \( \frac{1}{5} \)**: \( R_2 \leftarrow \frac{1}{5} R_2 \)
\[ \left( \begin{array}{ccc|ccc} 1 & 1 & -1 & -1 & 0 & 1 \\ 0 & 0 & 1 & \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \\ 0 & 0 & 10 & 5 & 1 & -4 \end{array} \right) \]

7. **Perform the row operation**: \( R_3 \leftarrow R_3 - 10R_2 \)
\[ \left( \begin{array}{ccc|ccc} 1 & 1 & -1 & -1 & 0 & 1 \\ 0 & 0 & 1 & \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \\ 0 & 0 & 0 & 0 & -1 & 0 \end{array} \right) \]

8. **Perform the row operation**: \( R_1 \leftarrow R_1 + R_2 \) and \( R_2 \leftarrow -R_2 \)
\[ \left( \begin{array}{ccc|ccc} 1 & 1 & 0 & -\frac{3}{5} & -\frac{1}{5} & \frac{3}{5} \\ 0 & 0 & 1 & \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array} \right) \]

9. **Final adjustments**: Perform necessary row operations to get the identity matrix on the left side.
\[ \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & -\frac{2}{5} & 0 & \frac{3}{5} \\ 0 & 1 & 0 & -\frac{1}{5} & \frac{1}{5} & 0 \\ 0 & 0 & 1 & \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \end{array} \right) \]

10. **Extract the inverse matrix**:
\[ A^{-1} = \begin{pmatrix} -\frac{2}{5} & 0 & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \end{pmatrix} \]