Solve the following cquations by matrix method :
(i) 3x+y=10
x+2y=5
(ii) x+3y=11
3x-y=3.

Answer

To solve the given equations using the matrix method, we will follow these steps for both sets of equations.

### Part (i): Solve the equations 3x + y = 10 and x + 2y = 5

1. **Write the equations in matrix form**:
The equations can be represented in the form \( AX = B \), where:
\[ A = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 10 \\ 5 \end{pmatrix} \]

2. **Find the inverse of matrix A**:
The inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:
\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
Here, \( a = 3, b = 1, c = 1, d = 2 \).
- Calculate the determinant \( ad - bc = 3 \cdot 2 - 1 \cdot 1 = 6 - 1 = 5 \).
- Thus, the inverse is:
\[ A^{-1} = \frac{1}{5} \begin{pmatrix} 2 & -1 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} \frac{2}{5} & -\frac{1}{5} \\ -\frac{1}{5} & \frac{3}{5} \end{pmatrix} \]

3. **Multiply A inverse with B**:
Now, we calculate \( X = A^{-1}B \):
\[ X = \begin{pmatrix} \frac{2}{5} & -\frac{1}{5} \\ -\frac{1}{5} & \frac{3}{5} \end{pmatrix} \begin{pmatrix} 10 \\ 5 \end{pmatrix} \]
- First row: \( \frac{2}{5} \cdot 10 + -\frac{1}{5} \cdot 5 = \frac{20}{5} - \frac{5}{5} = 4 - 1 = 3 \)
- Second row: \( -\frac{1}{5} \cdot 10 + \frac{3}{5} \cdot 5 = -\frac{10}{5} + \frac{15}{5} = -2 + 3 = 1 \)
- Thus, \( X = \begin{pmatrix} 3 \\ 1 \end{pmatrix} \).

4. **Solution**:
Therefore, the solution is \( x = 3 \) and \( y = 1 \).

### Part (ii): Solve the equations x + 3y = 11 and 3x - y = 3

1. **Write the equations in matrix form**:
The equations can be represented as:
\[ A = \begin{pmatrix} 1 & 3 \\ 3 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 11 \\ 3 \end{pmatrix} \]

2. **Find the inverse of matrix A**:
Calculate the determinant:
\[ ad - bc = 1 \cdot (-1) - 3 \cdot 3 = -1 - 9 = -10 \]
Thus, the inverse is:
\[ A^{-1} = \frac{1}{-10} \begin{pmatrix} -1 & -3 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{10} & \frac{3}{10} \\ \frac{3}{10} & -\frac{1}{10} \end{pmatrix} \]

3. **Multiply A inverse with B**:
Now, we calculate \( X = A^{-1}B \):
\[ X = \begin{pmatrix} \frac{1}{10} & \frac{3}{10} \\ \frac{3}{10} & -\frac{1}{10} \end{pmatrix} \begin{pmatrix} 11 \\ 3 \end{pmatrix} \]
- First row: \( \frac{1}{10} \cdot 11 + \frac{3}{10} \cdot 3 = \frac{11}{10} + \frac{9}{10} = \frac{20}{10} = 2 \)
- Second row: \( \frac{3}{10} \cdot 11 - \frac{1}{10} \cdot 3 = \frac{33}{10} - \frac{3}{10} = \frac{30}{10} = 3 \)
- Thus, \( X = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \).

4. **Solution**:
Therefore, the solution is \( x = 2 \) and \( y = 3 \).