`[{:(3,10),(2,7):}]` find the inverse of matrix

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} 3 & 10 \\ 2 & 7 \end{pmatrix} \), we will use the method of row operations to transform the matrix into the identity matrix while applying the same operations to an identity matrix.

### Step-by-step Solution:

1. **Set Up the Augmented Matrix**:
We start with the augmented matrix \( [A | I] \):
\[ \begin{pmatrix} 3 & 10 & | & 1 & 0 \\ 2 & 7 & | & 0 & 1 \end{pmatrix} \]

2. **Row Operation to Get Leading 1 in Row 1**:
We want to make the leading coefficient of the first row a 1. We can do this by dividing the entire first row by 3:
\[ R_1 = \frac{1}{3} R_1 \implies \begin{pmatrix} 1 & \frac{10}{3} & | & \frac{1}{3} & 0 \\ 2 & 7 & | & 0 & 1 \end{pmatrix} \]

3. **Eliminate the First Column of Row 2**:
Now we will eliminate the first column of the second row. We can do this by replacing \( R_2 \) with \( R_2 - 2R_1 \):
\[ R_2 = R_2 - 2R_1 \implies \begin{pmatrix} 1 & \frac{10}{3} & | & \frac{1}{3} & 0 \\ 0 & \frac{1}{3} & | & -\frac{2}{3} & 1 \end{pmatrix} \]

4. **Row Operation to Get Leading 1 in Row 2**:
Now we need to make the leading coefficient of the second row a 1. We can do this by multiplying the second row by 3:
\[ R_2 = 3R_2 \implies \begin{pmatrix} 1 & \frac{10}{3} & | & \frac{1}{3} & 0 \\ 0 & 1 & | & -2 & 3 \end{pmatrix} \]

5. **Eliminate the Second Column of Row 1**:
Now we will eliminate the second column of the first row. We can do this by replacing \( R_1 \) with \( R_1 - \frac{10}{3}R_2 \):
\[ R_1 = R_1 - \frac{10}{3}R_2 \implies \begin{pmatrix} 1 & 0 & | & 7 & -10 \\ 0 & 1 & | & -2 & 3 \end{pmatrix} \]

6. **Final Result**:
The left side of the augmented matrix is now the identity matrix, and the right side gives us the inverse of \( A \):
\[ A^{-1} = \begin{pmatrix} 7 & -10 \\ -2 & 3 \end{pmatrix} \]

### Final Answer:
The inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} 7 & -10 \\ -2 & 3 \end{pmatrix} \]