`[{:(3,1),(5,2):}]` using elementry method find the inverse of matrix

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \) using the elementary method, we will augment the matrix \( A \) with the identity matrix \( I \) and perform row operations to convert \( A \) into \( I \). The augmented matrix will look like this:

\[ \left( \begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 5 & 2 & 0 & 1 \end{array} \right) \]

### Step 1: Make the leading coefficient of the first row equal to 1
To do this, we divide the first row by 3:

\[ R_1 = \frac{1}{3} R_1 \]

This gives us:

\[ \left( \begin{array}{cc|cc} 1 & \frac{1}{3} & \frac{1}{3} & 0 \\ 5 & 2 & 0 & 1 \end{array} \right) \]

### Step 2: Eliminate the first element of the second row
Next, we want to make the first element of the second row equal to 0. We can do this by replacing the second row with \( R_2 - 5R_1 \):

\[ R_2 = R_2 - 5R_1 \]

Calculating this gives:

\[ R_2 = \begin{pmatrix} 5 & 2 & 0 & 1 \end{pmatrix} - 5 \begin{pmatrix} 1 & \frac{1}{3} & \frac{1}{3} & 0 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{3} & -\frac{5}{3} & 1 \end{pmatrix} \]

So the augmented matrix now looks like:

\[ \left( \begin{array}{cc|cc} 1 & \frac{1}{3} & \frac{1}{3} & 0 \\ 0 & \frac{1}{3} & -\frac{5}{3} & 1 \end{array} \right) \]

### Step 3: Make the leading coefficient of the second row equal to 1
Now, we will multiply the second row by 3:

\[ R_2 = 3R_2 \]

This gives us:

\[ \left( \begin{array}{cc|cc} 1 & \frac{1}{3} & \frac{1}{3} & 0 \\ 0 & 1 & -5 & 3 \end{array} \right) \]

### Step 4: Eliminate the second element of the first row
We need to make the second element of the first row equal to 0. We can do this by replacing the first row with \( R_1 - \frac{1}{3}R_2 \):

\[ R_1 = R_1 - \frac{1}{3}R_2 \]

Calculating this gives:

\[ R_1 = \begin{pmatrix} 1 & \frac{1}{3} & \frac{1}{3} & 0 \end{pmatrix} - \frac{1}{3} \begin{pmatrix} 0 & 1 & -5 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 2 & -1 \end{pmatrix} \]

So the augmented matrix now looks like:

\[ \left( \begin{array}{cc|cc} 1 & 0 & 2 & -1 \\ 0 & 1 & -5 & 3 \end{array} \right) \]

### Conclusion
The left side of the augmented matrix is now the identity matrix, and the right side gives us the inverse of the original matrix \( A \):

\[ A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \]