Find the inverse of the following , if it exists, by using elementary row (column) transformations :
`[(3,10),(2,7)]`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 3 & 10 \\ 2 & 7 \end{pmatrix} \) 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 of the same order (2x2):

\[ \begin{pmatrix} 3 & 10 & | & 1 & 0 \\ 2 & 7 & | & 0 & 1 \end{pmatrix} \]

### Step 2: Calculate the determinant
First, we need to check if the inverse exists by calculating the determinant of \(A\):

\[ \text{det}(A) = (3)(7) - (2)(10) = 21 - 20 = 1 \]

Since the determinant is not equal to zero, the inverse exists.

### Step 3: 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:

\[ R_1 \rightarrow \frac{1}{3} R_1 \Rightarrow \begin{pmatrix} 1 & \frac{10}{3} & | & \frac{1}{3} & 0 \\ 2 & 7 & | & 0 & 1 \end{pmatrix} \]

### Step 4: Eliminate the first column of the second row
Next, we want to make the first column of the second row equal to 0. We can do this by replacing the second row with \(R_2 - 2R_1\):

\[ R_2 \rightarrow R_2 - 2R_1 \Rightarrow \begin{pmatrix} 1 & \frac{10}{3} & | & \frac{1}{3} & 0 \\ 0 & \frac{1}{3} & | & -\frac{2}{3} & 1 \end{pmatrix} \]

### Step 5: 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 3:

\[ R_2 \rightarrow 3R_2 \Rightarrow \begin{pmatrix} 1 & \frac{10}{3} & | & \frac{1}{3} & 0 \\ 0 & 1 & | & -2 & 3 \end{pmatrix} \]

### Step 6: Eliminate the second column of the first row
Finally, we need to eliminate the second column of the first row. We can do this by replacing the first row with \(R_1 - \frac{10}{3} R_2\):

\[ R_1 \rightarrow R_1 - \frac{10}{3} R_2 \Rightarrow \begin{pmatrix} 1 & 0 & | & 7 & -10 \\ 0 & 1 & | & -2 & 3 \end{pmatrix} \]

### Step 7: Write the inverse matrix
Now, we have transformed the left side into the identity matrix. The right side gives us the inverse of \(A\):

\[ A^{-1} = \begin{pmatrix} 7 & -10 \\ -2 & 3 \end{pmatrix} \]

### Final Answer
The inverse of the matrix \( A = \begin{pmatrix} 3 & 10 \\ 2 & 7 \end{pmatrix} \) is:

\[ A^{-1} = \begin{pmatrix} 7 & -10 \\ -2 & 3 \end{pmatrix} \]