`[{:(2,-6),( 1,-2):}]` find inverse of matrix

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} 2 & -6 \\ 1 & -2 \end{pmatrix} \), we will use the method of elementary row operations. The inverse of a matrix \( A \) is denoted as \( A^{-1} \), and it satisfies the equation \( A A^{-1} = I \), where \( I \) is the identity matrix.

### Step-by-Step Solution:

1. **Set up the augmented matrix**:
We start by setting up the augmented matrix \([A | I]\):
\[ \begin{pmatrix} 2 & -6 & | & 1 & 0 \\ 1 & -2 & | & 0 & 1 \end{pmatrix} \]

2. **Make the leading coefficient of the first row equal to 1**:
We can achieve this by dividing the first row by 2:
\[ R_1 \leftarrow \frac{1}{2} R_1 \]
This gives us:
\[ \begin{pmatrix} 1 & -3 & | & \frac{1}{2} & 0 \\ 1 & -2 & | & 0 & 1 \end{pmatrix} \]

3. **Eliminate the first element of the second row**:
We subtract the first row from the second row:
\[ R_2 \leftarrow R_2 - R_1 \]
This results in:
\[ \begin{pmatrix} 1 & -3 & | & \frac{1}{2} & 0 \\ 0 & 1 & | & -\frac{1}{2} & 1 \end{pmatrix} \]

4. **Make the leading coefficient of the second row equal to 1**:
The leading coefficient of the second row is already 1, so we can proceed to eliminate the second element of the first row:
\[ R_1 \leftarrow R_1 + 3R_2 \]
This gives us:
\[ \begin{pmatrix} 1 & 0 & | & \frac{1}{2} + 3(-\frac{1}{2}) & 3 \\ 0 & 1 & | & -\frac{1}{2} & 1 \end{pmatrix} \]
Simplifying the first row:
\[ \begin{pmatrix} 1 & 0 & | & -\frac{1}{2} & 3 \\ 0 & 1 & | & -\frac{1}{2} & 1 \end{pmatrix} \]

5. **Final adjustments**:
We can express the inverse matrix from the augmented matrix:
\[ A^{-1} = \begin{pmatrix} -\frac{1}{2} & 3 \\ -\frac{1}{2} & 1 \end{pmatrix} \]

### Final Answer:
The inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} -\frac{1}{2} & 3 \\ -\frac{1}{2} & 1 \end{pmatrix} \]