`[{:(2,1),(1,1):}]` find the inverse of matrix

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \), we will use the method of row operations to convert the matrix into the identity matrix while applying the same operations to the identity matrix.

### Step-by-Step Solution:

1. **Set Up the Augmented Matrix**:
We start with the augmented matrix \([A | I]\), where \(I\) is the identity matrix:
\[ \left( \begin{array}{cc|cc} 2 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array} \right) \]

2. **Make the First Pivot 1**:
To make the first element of the first row (the pivot) equal to 1, we can divide the first row by 2:
\[ R_1 = \frac{1}{2} R_1 \]
This gives us:
\[ \left( \begin{array}{cc|cc} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 1 & 1 & 0 & 1 \end{array} \right) \]

3. **Eliminate the First Column of the Second Row**:
Now, we need to make the first element of the second row equal to 0. We can do this by subtracting the first row from the second row:
\[ R_2 = R_2 - R_1 \]
This results in:
\[ \left( \begin{array}{cc|cc} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 1 \end{array} \right) \]

4. **Make the Second Pivot 1**:
Next, we make the pivot in the second row equal to 1 by multiplying the second row by 2:
\[ R_2 = 2 R_2 \]
This gives us:
\[ \left( \begin{array}{cc|cc} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 1 & -1 & 2 \end{array} \right) \]

5. **Eliminate the Second Column of the First Row**:
Finally, we need to eliminate the second column of the first row. We do this by subtracting \(\frac{1}{2}\) times the second row from the first row:
\[ R_1 = R_1 - \frac{1}{2} R_2 \]
This results in:
\[ \left( \begin{array}{cc|cc} 1 & 0 & 1 & -1 \\ 0 & 1 & -1 & 2 \end{array} \right) \]

6. **Read the Inverse Matrix**:
The left side of the augmented matrix is now the identity matrix, and the right side gives us the inverse of \(A\):
\[ A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} \]

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