Find `A^(-1)` using column transformations :
`A = [(2,-3),(-1,2)]`

Answer

To find the inverse of the matrix \( A \) using column transformations, we start with the matrix \( A \) and the identity matrix \( I \).

Given:
\[ A = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} \]
and the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

We will perform column operations on \( A \) to transform it into \( I \) while applying the same operations to \( I \).

### Step 1: Set up the augmented matrix
We write the augmented matrix \([A | I]\):
\[ \begin{pmatrix} 2 & -3 & | & 1 & 0 \\ -1 & 2 & | & 0 & 1 \end{pmatrix} \]

### Step 2: Make the first column's leading entry 1
To make the leading entry of the first column 1, we can divide the first row by 2:
\[ R_1 \rightarrow \frac{1}{2} R_1 \]
This gives us:
\[ \begin{pmatrix} 1 & -\frac{3}{2} & | & \frac{1}{2} & 0 \\ -1 & 2 & | & 0 & 1 \end{pmatrix} \]

### Step 3: Eliminate the entry below the leading 1
Next, we want to eliminate the entry below the leading 1 in the first column. We can do this by adding the first row to the second row:
\[ R_2 \rightarrow R_2 + R_1 \]
This results in:
\[ \begin{pmatrix} 1 & -\frac{3}{2} & | & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & | & \frac{1}{2} & 1 \end{pmatrix} \]

### Step 4: Make the second column's leading entry 1
Now, we need to make the leading entry of the second column 1 by multiplying the second row by 2:
\[ R_2 \rightarrow 2R_2 \]
This gives us:
\[ \begin{pmatrix} 1 & -\frac{3}{2} & | & \frac{1}{2} & 0 \\ 0 & 1 & | & 1 & 2 \end{pmatrix} \]

### Step 5: Eliminate the entry above the leading 1 in the second column
Now we need to eliminate the entry above the leading 1 in the second column. We can do this by adding \(\frac{3}{2}\) times the second row to the first row:
\[ R_1 \rightarrow R_1 + \frac{3}{2} R_2 \]
This results in:
\[ \begin{pmatrix} 1 & 0 & | & 2 & 3 \\ 0 & 1 & | & 1 & 2 \end{pmatrix} \]

### Step 6: Write the inverse
Now, the left side is the identity matrix, and the right side gives us the inverse of \( A \):
\[ A^{-1} = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \]

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