Using elementary transformation, find the inverse of the matrix `({:(2, -1, 1), (-1, 2, -1), (1, -1, 2):})`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} \) using elementary transformations, we will augment the matrix \( A \) with the identity matrix \( I \) and perform row operations until the left side becomes the identity matrix. The right side will then be the inverse of \( A \).

### Step-by-Step Solution:

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

2. **First transformation**: \( R_1 \leftarrow R_1 + R_2 \) and \( R_2 \leftarrow R_2 + R_3 \):
\[ \left( \begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 1 \\ 1 & -1 & 2 & 0 & 0 & 1 \end{array} \right) \]

3. **Second transformation**: \( R_3 \leftarrow R_3 - R_1 \):
\[ \left( \begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 1 \\ 0 & -2 & 2 & -1 & -1 & 1 \end{array} \right) \]

4. **Third transformation**: \( R_3 \leftarrow R_3 + 2R_2 \):
\[ \left( \begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 1 \\ 0 & 0 & 4 & -1 & 1 & 3 \end{array} \right) \]

5. **Fourth transformation**: \( R_3 \leftarrow \frac{1}{4} R_3 \):
\[ \left( \begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & -\frac{1}{4} & \frac{1}{4} & \frac{3}{4} \end{array} \right) \]

6. **Fifth transformation**: \( R_2 \leftarrow R_2 - R_3 \) and \( R_1 \leftarrow R_1 - R_2 \):
\[ \left( \begin{array}{ccc|ccc} 1 & 0 & -1 & 1 & 0 & -1 \\ 0 & 1 & 0 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ 0 & 0 & 1 & -\frac{1}{4} & \frac{1}{4} & \frac{3}{4} \end{array} \right) \]

7. **Final transformation**: \( R_1 \leftarrow R_1 + R_3 \):
\[ \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\ 0 & 1 & 0 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ 0 & 0 & 1 & -\frac{1}{4} & \frac{1}{4} & \frac{3}{4} \end{array} \right) \]

Now the left side is the identity matrix, and the right side gives us the inverse of \( A \):
\[ A^{-1} = \begin{pmatrix} \frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ -\frac{1}{4} & \frac{1}{4} & \frac{3}{4} \end{pmatrix} \]