Using elementary transformations, find the inverse of,
`[(5,2),(2,1)]`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 5 & 2 \\ 2 & 1 \end{pmatrix} \) using elementary transformations, we will augment the matrix \( A \) with the identity matrix \( I \) and apply row operations to transform \( A \) into \( I \). The augmented matrix will look like this:

\[ \left( \begin{array}{cc|cc} 5 & 2 & 1 & 0 \\ 2 & 1 & 0 & 1 \end{array} \right) \]

### Step 1: Make the leading coefficient of the first row equal to 1
We can achieve this by dividing the first row by 5:

\[ R_1 \rightarrow \frac{1}{5} R_1 \]

This gives us:

\[ \left( \begin{array}{cc|cc} 1 & \frac{2}{5} & \frac{1}{5} & 0 \\ 2 & 1 & 0 & 1 \end{array} \right) \]

### Step 2: Eliminate the first element of the second row
Next, we will eliminate the first element of the second row by performing the operation:

\[ R_2 \rightarrow R_2 - 2R_1 \]

Calculating this gives:

\[ R_2 = 2 - 2(1) = 0 \\ R_2 = 1 - 2\left(\frac{2}{5}\right) = 1 - \frac{4}{5} = \frac{1}{5} \\ R_2 = 0 - 2\left(\frac{1}{5}\right) = -\frac{2}{5} \]

So, the augmented matrix now looks like:

\[ \left( \begin{array}{cc|cc} 1 & \frac{2}{5} & \frac{1}{5} & 0 \\ 0 & \frac{1}{5} & -\frac{2}{5} & 1 \end{array} \right) \]

### Step 3: Make the leading coefficient of the second row equal to 1
We can achieve this by multiplying the second row by 5:

\[ R_2 \rightarrow 5R_2 \]

This gives us:

\[ \left( \begin{array}{cc|cc} 1 & \frac{2}{5} & \frac{1}{5} & 0 \\ 0 & 1 & -2 & 5 \end{array} \right) \]

### Step 4: Eliminate the second element of the first row
Now we will eliminate the second element of the first row by performing the operation:

\[ R_1 \rightarrow R_1 - \frac{2}{5}R_2 \]

Calculating this gives:

\[ R_1 = 1 - \frac{2}{5}(0) = 1 \\ R_1 = \frac{2}{5} - \frac{2}{5}(1) = 0 \\ R_1 = \frac{1}{5} - \frac{2}{5}(-2) = \frac{1}{5} + \frac{4}{5} = 1 \]

So, the augmented matrix now looks like:

\[ \left( \begin{array}{cc|cc} 1 & 0 & 5 & -2 \\ 0 & 1 & -2 & 5 \end{array} \right) \]

### Conclusion
Now we have transformed the left side into the identity matrix. The right side of the augmented matrix gives us the inverse of the original matrix \( A \):

\[ A^{-1} = \begin{pmatrix} 5 & -2 \\ -2 & 5 \end{pmatrix} \]