By using elementary transformations, find the inverse of : `A=[(3,1),(5,2)]`.

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \) using elementary transformations, we will follow these steps:

### Step 1: Set up the augmented matrix
We start by creating an augmented matrix that combines \( A \) with the identity matrix \( I \):
\[ \left( \begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 5 & 2 & 0 & 1 \end{array} \right) \]

### Step 2: Apply Row Operation R2 → R2 - R1
We will perform the row operation \( R_2 \rightarrow R_2 - R_1 \):
\[ R_2 = R_2 - R_1 \implies \left( \begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 5 - 3 & 2 - 1 & 0 - 1 & 1 - 0 \end{array} \right) = \left( \begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 2 & 1 & -1 & 1 \end{array} \right) \]

### Step 3: Apply Row Operation R1 → R1 - R2
Now we will perform the operation \( R_1 \rightarrow R_1 - R_2 \):
\[ R_1 = R_1 - R_2 \implies \left( \begin{array}{cc|cc} 3 - 2 & 1 - 1 & 1 - (-1) & 0 - 1 \\ 2 & 1 & -1 & 1 \end{array} \right) = \left( \begin{array}{cc|cc} 1 & 0 & 2 & -1 \\ 2 & 1 & -1 & 1 \end{array} \right) \]

### Step 4: Apply Row Operation R2 → R2 - 2R1
Next, we will perform the operation \( R_2 \rightarrow R_2 - 2R_1 \):
\[ R_2 = R_2 - 2R_1 \implies \left( \begin{array}{cc|cc} 1 & 0 & 2 & -1 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot 0 & -1 - 2 \cdot 2 & 1 - 2 \cdot (-1) \end{array} \right) = \left( \begin{array}{cc|cc} 1 & 0 & 2 & -1 \\ 0 & 1 & -5 & 3 \end{array} \right) \]

### Step 5: The left side is now the identity matrix
Now we have the augmented matrix in the form:
\[ \left( \begin{array}{cc|cc} 1 & 0 & 2 & -1 \\ 0 & 1 & -5 & 3 \end{array} \right) \]
This indicates that the inverse of \( A \) is:
\[ A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \]

### Final Answer
Thus, the inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \]