`[{:(1,3),(2,7):}]` find the inverse of matrix

Explanation

To find the inverse of the matrix \(\begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix}\), we will follow these steps:

### Step 1: Define the Matrix
Let \( A = \begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix} \).

### Step 2: Calculate the Determinant
The determinant of a \(2 \times 2\) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by the formula:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A \):
- \( a = 1 \)
- \( b = 3 \)
- \( c = 2 \)
- \( d = 7 \)

Calculating the determinant:
\[ \text{det}(A) = (1)(7) - (2)(3) = 7 - 6 = 1 \]

### Step 3: Check if the Determinant is Non-Zero
Since the determinant is \( 1 \) (which is not zero), the inverse exists.

### Step 4: Calculate the Adjoint
To find the adjoint of matrix \( A \), we need to calculate the cofactors.

1. **Calculate Minors:**
- \( M_{11} = 7 \) (remove first row and first column)
- \( M_{12} = 2 \) (remove first row and second column)
- \( M_{21} = 3 \) (remove second row and first column)
- \( M_{22} = 1 \) (remove second row and second column)

2. **Calculate Cofactors:**
- \( C_{11} = (-1)^{1+1} \cdot M_{11} = 1 \cdot 7 = 7 \)
- \( C_{12} = (-1)^{1+2} \cdot M_{12} = -1 \cdot 2 = -2 \)
- \( C_{21} = (-1)^{2+1} \cdot M_{21} = -1 \cdot 3 = -3 \)
- \( C_{22} = (-1)^{2+2} \cdot M_{22} = 1 \cdot 1 = 1 \)

Thus, the cofactor matrix is:
\[ \text{Cofactor}(A) = \begin{pmatrix} 7 & -2 \\ -3 & 1 \end{pmatrix} \]

3. **Transpose the Cofactor Matrix to get the Adjoint:**
\[ \text{Adj}(A) = \text{Cofactor}(A)^T = \begin{pmatrix} 7 & -3 \\ -2 & 1 \end{pmatrix} \]

### Step 5: Calculate the Inverse
The inverse of matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adj}(A) \]
Substituting the values we found:
\[ A^{-1} = \frac{1}{1} \cdot \begin{pmatrix} 7 & -3 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 7 & -3 \\ -2 & 1 \end{pmatrix} \]

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