Find the inverse of the following , if it exists, by using elementary row (column) transformations :
`[(2,-3),(-1,2)]`

Answer

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

### Step 1: Write the augmented matrix
We start by writing the augmented matrix \([A | I]\), where \(I\) is the identity matrix:

\[ \begin{pmatrix} 2 & -3 & | & 1 & 0 \\ -1 & 2 & | & 0 & 1 \end{pmatrix} \]

### Step 2: Apply row operations to get the identity matrix on the left side
We will perform row operations to transform the left side into the identity matrix.

**Operation 1:** Multiply the first row by \(\frac{1}{2}\):

\[ R_1 \rightarrow \frac{1}{2} R_1 \]

This gives us:

\[ \begin{pmatrix} 1 & -\frac{3}{2} & | & \frac{1}{2} & 0 \\ -1 & 2 & | & 0 & 1 \end{pmatrix} \]

**Operation 2:** Add the first row to the second row:

\[ R_2 \rightarrow R_2 + R_1 \]

This gives us:

\[ \begin{pmatrix} 1 & -\frac{3}{2} & | & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & | & \frac{1}{2} & 1 \end{pmatrix} \]

**Operation 3:** Multiply the second row by \(2\):

\[ R_2 \rightarrow 2 R_2 \]

This gives us:

\[ \begin{pmatrix} 1 & -\frac{3}{2} & | & \frac{1}{2} & 0 \\ 0 & 1 & | & 1 & 2 \end{pmatrix} \]

**Operation 4:** Add \(\frac{3}{2}\) times the second row to the first row:

\[ R_1 \rightarrow R_1 + \frac{3}{2} R_2 \]

This gives us:

\[ \begin{pmatrix} 1 & 0 & | & 2 & 3 \\ 0 & 1 & | & 1 & 2 \end{pmatrix} \]

### Step 3: Read off the inverse
Now that the left side is the identity matrix, the right side gives us the inverse of \(A\):

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

### Conclusion
Thus, the inverse of the matrix \( A = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} \) is

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