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

Explanation

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

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

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

Calculating the determinant:
\[ \text{det}(A) = (1)(3) - (-1)(2) = 3 + 2 = 5 \]

### Step 3: Verify the Determinant
Since \(\text{det}(A) = 5\) which is not equal to 0, the inverse of the matrix exists.

### Step 4: Find the Adjoint of the Matrix
The adjoint of a \(2 \times 2\) matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is given by:
\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
For our matrix:
\[ \text{adj}(A) = \begin{pmatrix} 3 & 1 \\ -2 & 1 \end{pmatrix} \]

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

### Final Answer
The inverse of the matrix \(\begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}\) is:
\[ A^{-1} = \begin{pmatrix} \frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} \]