Solve the equations by matrix method : (i) x+2y+z=7 x+2y+z=7 x+3z=11

Answer

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

### Given Equations:
1. \( x + 2y + z = 7 \) (Equation 1)
2. \( x + 2y + z = 7 \) (Equation 2)
3. \( x + 3z = 11 \) (Equation 3)

### Step 1: Write the equations in matrix form
We can represent the system of equations in the form \( AX = B \), where:
- \( A \) is the coefficient matrix,
- \( X \) is the variable matrix,
- \( B \) is the constant matrix.

From the equations, we can identify:
\[ A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ 7 \\ 11 \end{bmatrix} \]

### Step 2: Calculate the determinant of matrix \( A \)
To check if the inverse of matrix \( A \) exists, we need to calculate its determinant \( |A| \).

\[ |A| = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the remaining rows.

Calculating:
\[ |A| = 1 \cdot (2 \cdot 3 - 1 \cdot 0) - 2 \cdot (1 \cdot 3 - 1 \cdot 1) + 1 \cdot (1 \cdot 0 - 2 \cdot 1) \]
\[ = 1 \cdot (6) - 2 \cdot (3 - 1) + 1 \cdot (0 - 2) \]
\[ = 6 - 2 \cdot 2 - 2 \]
\[ = 6 - 4 - 2 = 0 \]

### Step 3: Analyze the determinant
Since the determinant \( |A| = 0 \), this indicates that the matrix \( A \) is singular, which means that the system of equations does not have a unique solution.

### Conclusion
The system of equations is inconsistent, and therefore, there are no solutions.