Solve the following equations by inverse method :
`2x +5y = 1, 3x + 2y = 7`

Answer

To solve the equations \(2x + 5y = 1\) and \(3x + 2y = 7\) using the inverse method, we will follow these steps:

### Step 1: Write the equations in matrix form
We can represent the system of equations in matrix form as follows:

\[ \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 7 \end{bmatrix} \]

Let \(A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \end{bmatrix}\), and \(B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}\).

### Step 2: Find the inverse of matrix A
To find \(X\), we need to calculate \(A^{-1}\). The formula for the inverse of a 2x2 matrix is given by:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

#### Step 2.1: Calculate the determinant of A
The determinant of \(A\) is calculated as follows:

\[ \text{det}(A) = (2)(2) - (5)(3) = 4 - 15 = -11 \]

#### Step 2.2: Calculate the adjoint of A
The adjoint of \(A\) is obtained by swapping the elements on the main diagonal and changing the signs of the off-diagonal elements:

\[ \text{adj}(A) = \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix} \]

#### Step 2.3: Calculate the inverse of A
Now we can find \(A^{-1}\):

\[ A^{-1} = \frac{1}{-11} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{2}{11} & \frac{5}{11} \\ \frac{3}{11} & -\frac{2}{11} \end{bmatrix} \]

### Step 3: Multiply \(A^{-1}\) by B to find X
Now we can find \(X\) by multiplying \(A^{-1}\) with \(B\):

\[ X = A^{-1}B = \begin{bmatrix} -\frac{2}{11} & \frac{5}{11} \\ \frac{3}{11} & -\frac{2}{11} \end{bmatrix} \begin{bmatrix} 1 \\ 7 \end{bmatrix} \]

Calculating the multiplication:

\[ X = \begin{bmatrix} -\frac{2}{11} \cdot 1 + \frac{5}{11} \cdot 7 \\ \frac{3}{11} \cdot 1 + -\frac{2}{11} \cdot 7 \end{bmatrix} = \begin{bmatrix} -\frac{2}{11} + \frac{35}{11} \\ \frac{3}{11} - \frac{14}{11} \end{bmatrix} = \begin{bmatrix} \frac{33}{11} \\ -\frac{11}{11} \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix} \]

### Conclusion
Thus, the solution to the equations is:

\[ x = 3, \quad y = -1 \]