Find the inverase of A`=[{:(5,-1),(1,1):}]`by using elementary row transformation.

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} 5 & -1 \\ 1 & 1 \end{pmatrix} \) using elementary row transformations, we will augment the matrix \( A \) with the identity matrix and perform row operations until we obtain the identity matrix on the left side. The resulting matrix on the right side will be the inverse of \( A \).

### Step-by-Step Solution:

1. **Set Up the Augmented Matrix**:
We start with the augmented matrix \( [A | I] \):
\[ \begin{pmatrix} 5 & -1 & | & 1 & 0 \\ 1 & 1 & | & 0 & 1 \end{pmatrix} \]

2. **Make the Leading Coefficient of the First Row a 1**:
We can do this by dividing the first row by 5:
\[ R_1 \leftarrow \frac{1}{5} R_1 \]
This gives us:
\[ \begin{pmatrix} 1 & -\frac{1}{5} & | & \frac{1}{5} & 0 \\ 1 & 1 & | & 0 & 1 \end{pmatrix} \]

3. **Eliminate the First Element of the Second Row**:
We subtract the first row from the second row to eliminate the leading 1 in the second row:
\[ R_2 \leftarrow R_2 - R_1 \]
This results in:
\[ \begin{pmatrix} 1 & -\frac{1}{5} & | & \frac{1}{5} & 0 \\ 0 & 1 + \frac{1}{5} & | & -\frac{1}{5} & 1 \end{pmatrix} \]
Simplifying the second row gives:
\[ \begin{pmatrix} 1 & -\frac{1}{5} & | & \frac{1}{5} & 0 \\ 0 & \frac{6}{5} & | & -\frac{1}{5} & 1 \end{pmatrix} \]

4. **Make the Leading Coefficient of the Second Row a 1**:
We can achieve this by multiplying the second row by \( \frac{5}{6} \):
\[ R_2 \leftarrow \frac{5}{6} R_2 \]
This gives us:
\[ \begin{pmatrix} 1 & -\frac{1}{5} & | & \frac{1}{5} & 0 \\ 0 & 1 & | & -\frac{5}{36} & \frac{5}{6} \end{pmatrix} \]

5. **Eliminate the Second Element of the First Row**:
We now eliminate the second element of the first row by adding \( \frac{1}{5} \) times the second row to the first row:
\[ R_1 \leftarrow R_1 + \frac{1}{5} R_2 \]
This results in:
\[ \begin{pmatrix} 1 & 0 & | & \frac{1}{5} - \frac{1}{5} \cdot \frac{5}{36} & \frac{1}{5} \cdot \frac{5}{6} \\ 0 & 1 & | & -\frac{5}{36} & \frac{5}{6} \end{pmatrix} \]
Simplifying gives:
\[ \begin{pmatrix} 1 & 0 & | & \frac{7}{36} & \frac{1}{6} \\ 0 & 1 & | & -\frac{5}{36} & \frac{5}{6} \end{pmatrix} \]

6. **Final Result**:
The right side of the augmented matrix now represents the inverse of \( A \):
\[ A^{-1} = \begin{pmatrix} \frac{7}{36} & \frac{1}{6} \\ -\frac{5}{36} & \frac{5}{6} \end{pmatrix} \]