If `A=[{:(2,-3),(-1,2):}]` then `A^(-1)=`

Options

A. `[{:(2,-3),( -1,2):}]`
B. `[{:(2,3),( 1,2):}]`
C. `[{:( -1,2),(2,-3):}]`
D. `[{:( 1,2),(2,3):}]`

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} \), we can use the method of row reduction to convert the augmented matrix \( [A | I] \) into the form \( [I | A^{-1}] \), where \( I \) is the identity matrix.

### Step-by-Step Solution:

1. **Set up the augmented matrix**:
\[ [A | I] = \left[ \begin{array}{cc|cc} 2 & -3 & 1 & 0 \\ -1 & 2 & 0 & 1 \end{array} \right] \]

2. **Make the leading coefficient of the first row equal to 1**:
To do this, we can divide the first row by 2:
\[ R_1 \leftarrow \frac{1}{2} R_1 \Rightarrow \left[ \begin{array}{cc|cc} 1 & -\frac{3}{2} & \frac{1}{2} & 0 \\ -1 & 2 & 0 & 1 \end{array} \right] \]

3. **Eliminate the first column of the second row**:
We can add the first row to the second row:
\[ R_2 \leftarrow R_2 + R_1 \Rightarrow \left[ \begin{array}{cc|cc} 1 & -\frac{3}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \end{array} \right] \]

4. **Make the leading coefficient of the second row equal to 1**:
To do this, we can multiply the second row by 2:
\[ R_2 \leftarrow 2 R_2 \Rightarrow \left[ \begin{array}{cc|cc} 1 & -\frac{3}{2} & \frac{1}{2} & 0 \\ 0 & 1 & 1 & 2 \end{array} \right] \]

5. **Eliminate the second column of the first row**:
We can add \(\frac{3}{2}\) times the second row to the first row:
\[ R_1 \leftarrow R_1 + \frac{3}{2} R_2 \Rightarrow \left[ \begin{array}{cc|cc} 1 & 0 & 2 & 3 \\ 0 & 1 & 1 & 2 \end{array} \right] \]

6. **Final augmented matrix**:
The augmented matrix is now:
\[ \left[ \begin{array}{cc|cc} 1 & 0 & 2 & 3 \\ 0 & 1 & 1 & 2 \end{array} \right] \]
This means that:
\[ A^{-1} = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \]

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