Using elementary transformations, find the inverse of the matrix`[[3,-1],[-4, 2]]`

Explanation

To find the inverse of the matrix \( A = \begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix} \) using elementary transformations, we will augment the matrix \( A \) with the identity matrix \( I \) and perform row operations to transform \( A \) into \( I \).

### Step 1: Set up the augmented matrix
We start with the augmented matrix \( [A | I] \):
\[ \begin{bmatrix} 3 & -1 & | & 1 & 0 \\ -4 & 2 & | & 0 & 1 \end{bmatrix} \]

### Step 2: Make the leading coefficient of the first row equal to 1
To make the leading coefficient of the first row equal to 1, we can divide the first row by 3:
\[ R_1 \rightarrow \frac{1}{3} R_1 \]
This gives us:
\[ \begin{bmatrix} 1 & -\frac{1}{3} & | & \frac{1}{3} & 0 \\ -4 & 2 & | & 0 & 1 \end{bmatrix} \]

### Step 3: Eliminate the first column of the second row
Next, we want to eliminate the first column of the second row. We can do this by adding 4 times the first row to the second row:
\[ R_2 \rightarrow R_2 + 4R_1 \]
This results in:
\[ \begin{bmatrix} 1 & -\frac{1}{3} & | & \frac{1}{3} & 0 \\ 0 & \frac{2}{3} & | & \frac{4}{3} & 1 \end{bmatrix} \]

### Step 4: Make the leading coefficient of the second row equal to 1
Now, we need to make the leading coefficient of the second row equal to 1 by multiplying the second row by \( \frac{3}{2} \):
\[ R_2 \rightarrow \frac{3}{2} R_2 \]
This gives us:
\[ \begin{bmatrix} 1 & -\frac{1}{3} & | & \frac{1}{3} & 0 \\ 0 & 1 & | & 2 & \frac{3}{2} \end{bmatrix} \]

### Step 5: Eliminate the second column of the first row
Finally, we eliminate the second column of the first row by adding \( \frac{1}{3} \) times the second row to the first row:
\[ R_1 \rightarrow R_1 + \frac{1}{3} R_2 \]
This results in:
\[ \begin{bmatrix} 1 & 0 & | & 1 & \frac{1}{2} \\ 0 & 1 & | & 2 & \frac{3}{2} \end{bmatrix} \]

### Step 6: Read the inverse from the augmented matrix
The left side of the augmented matrix is now the identity matrix, and the right side gives us the inverse of \( A \):
\[ A^{-1} = \begin{bmatrix} 1 & \frac{1}{2} \\ 2 & \frac{3}{2} \end{bmatrix} \]

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