Using elementary transformations find the inverse of the matrix
`A=[(2,1),(4,7)]`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 2 & 1 \\ 4 & 7 \end{pmatrix} \) using elementary transformations, we will augment the matrix \( A \) with the identity matrix \( I \) and perform row operations to transform \( A \) into \( I \).

### Step 1: Set up the augmented matrix
We start with the augmented matrix \( [A | I] \):
\[ \begin{pmatrix} 2 & 1 & | & 1 & 0 \\ 4 & 7 & | & 0 & 1 \end{pmatrix} \]

### Step 2: Make the leading coefficient of the first row equal to 1
We can achieve this by dividing the first row by 2:
\[ R_1 \rightarrow \frac{1}{2} R_1 \]
This gives us:
\[ \begin{pmatrix} 1 & \frac{1}{2} & | & \frac{1}{2} & 0 \\ 4 & 7 & | & 0 & 1 \end{pmatrix} \]

### Step 3: Eliminate the first column of the second row
Next, we will eliminate the 4 in the second row by replacing \( R_2 \) with \( R_2 - 4R_1 \):
\[ R_2 \rightarrow R_2 - 4R_1 \]
Calculating this gives:
\[ R_2 = (4 - 4 \cdot 1, 7 - 4 \cdot \frac{1}{2}, 0 - 4 \cdot \frac{1}{2}, 1 - 4 \cdot 0) = (0, 5 - 2, -2, 1) \]
So we have:
\[ \begin{pmatrix} 1 & \frac{1}{2} & | & \frac{1}{2} & 0 \\ 0 & 5 & | & -2 & 1 \end{pmatrix} \]

### Step 4: Make the leading coefficient of the second row equal to 1
Now, we divide the second row by 5:
\[ R_2 \rightarrow \frac{1}{5} R_2 \]
This gives us:
\[ \begin{pmatrix} 1 & \frac{1}{2} & | & \frac{1}{2} & 0 \\ 0 & 1 & | & -\frac{2}{5} & \frac{1}{5} \end{pmatrix} \]

### Step 5: Eliminate the second column of the first row
Next, we eliminate the \( \frac{1}{2} \) in the first row by replacing \( R_1 \) with \( R_1 - \frac{1}{2} R_2 \):
\[ R_1 \rightarrow R_1 - \frac{1}{2} R_2 \]
Calculating this gives:
\[ R_1 = (1 - 0, \frac{1}{2} - \frac{1}{2}, \frac{1}{2} - \frac{1}{2} \cdot -\frac{2}{5}, 0 - \frac{1}{2} \cdot \frac{1}{5}) = (1, 0, \frac{1}{2} + \frac{1}{5}, -\frac{1}{10}) \]
So we have:
\[ \begin{pmatrix} 1 & 0 & | & \frac{5}{10} + \frac{2}{10} & -\frac{1}{10} \\ 0 & 1 & | & -\frac{2}{5} & \frac{1}{5} \end{pmatrix} \]
This simplifies to:
\[ \begin{pmatrix} 1 & 0 & | & \frac{7}{10} & -\frac{1}{10} \\ 0 & 1 & | & -\frac{2}{5} & \frac{1}{5} \end{pmatrix} \]

### Step 6: Write down the inverse
The right side of the augmented matrix now represents the inverse of \( A \):
\[ A^{-1} = \begin{pmatrix} \frac{7}{10} & -\frac{1}{10} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} \]

### Final Answer
Thus, the inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} \frac{7}{10} & -\frac{1}{10} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} \]