Solve the following system of equations by matrix method :
`3x-4y=5` and `4x+2y=3`

Answer

To solve the system of equations using the matrix method, we will follow these steps:

### Given Equations:
1. \( 3x - 4y = 5 \) (Equation 1)
2. \( 4x + 2y = 3 \) (Equation 2)

### Step 1: Write the equations in matrix form
We can express the system of equations in the form \( A \mathbf{x} = \mathbf{b} \), where:

\[ A = \begin{bmatrix} 3 & -4 \\ 4 & 2 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 5 \\ 3 \end{bmatrix} \]

### Step 2: Find the determinant of matrix A
The determinant of matrix \( A \) is calculated as follows:

\[ \text{det}(A) = (3)(2) - (-4)(4) = 6 + 16 = 22 \]

### Step 3: Find the adjoint of matrix A
To find the adjoint of matrix \( A \), we swap the elements on the main diagonal and change the signs of the off-diagonal elements:

\[ \text{adj}(A) = \begin{bmatrix} 2 & 4 \\ -4 & 3 \end{bmatrix} \]

### Step 4: Find the inverse of matrix A
The inverse of matrix \( A \) is given by:

\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = \frac{1}{22} \begin{bmatrix} 2 & 4 \\ -4 & 3 \end{bmatrix} \]

### Step 5: Multiply \( A^{-1} \) with \( \mathbf{b} \)
Now we will multiply \( A^{-1} \) with \( \mathbf{b} \):

\[ \mathbf{x} = A^{-1} \mathbf{b} = \frac{1}{22} \begin{bmatrix} 2 & 4 \\ -4 & 3 \end{bmatrix} \begin{bmatrix} 5 \\ 3 \end{bmatrix} \]

Calculating the multiplication:

\[ = \frac{1}{22} \begin{bmatrix} (2 \cdot 5 + 4 \cdot 3) \\ (-4 \cdot 5 + 3 \cdot 3) \end{bmatrix} = \frac{1}{22} \begin{bmatrix} 10 + 12 \\ -20 + 9 \end{bmatrix} = \frac{1}{22} \begin{bmatrix} 22 \\ -11 \end{bmatrix} \]

### Step 6: Simplify the results
Now we simplify the results:

\[ \mathbf{x} = \begin{bmatrix} 1 \\ -\frac{1}{2} \end{bmatrix} \]

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

\[ x = 1, \quad y = -\frac{1}{2} \]