The system of equations `x-y+3z=4,x+z=2` `x+y-z=0` has

Options

A. a unique solution
B. finitely many solutions
C. infinitely many solutions
D. none of these

Answer

To determine the nature of the solutions for the given system of equations:

1. **Equations:**
- \( x - y + 3z = 4 \) (Equation 1)
- \( x + z = 2 \) (Equation 2)
- \( x + y - z = 0 \) (Equation 3)

2. **Forming the Matrix:**
We can represent the system of equations in matrix form as \( AX = B \), where:
\[ A = \begin{bmatrix} 1 & -1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 4 \\ 2 \\ 0 \end{bmatrix} \]

3. **Finding the Determinant (Del):**
We need to calculate the determinant of matrix \( A \):
\[ \text{Del} = \begin{vmatrix} 1 & -1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]
Using the determinant formula for a 3x3 matrix:
\[ \text{Del} = 1 \cdot \begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} \]

Calculating the minors:
- \( \begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix} = (0)(-1) - (1)(1) = -1 \)
- \( \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = (1)(-1) - (1)(1) = -2 \)
- \( \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = (1)(1) - (0)(1) = 1 \)

Substituting these values back into the determinant:
\[ \text{Del} = 1 \cdot (-1) - (-1) \cdot (-2) + 3 \cdot 1 = -1 - 2 + 3 = 0 \]

4. **Finding Determinants for Infinite Solutions:**
Since \( \text{Del} = 0 \), we need to check the determinants of the modified matrices \( \text{Del}_1 \) and \( \text{Del}_2 \) to determine if there are infinitely many solutions.

- **Calculating \( \text{Del}_1 \)**:
\[ \text{Del}_1 = \begin{vmatrix} 4 & -1 & 3 \\ 2 & 0 & 1 \\ 0 & 1 & -1 \end{vmatrix} \]
Calculating this determinant:
\[ \text{Del}_1 = 4 \cdot \begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & 1 \\ 0 & -1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} \]
- The minors yield:
- \( \begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix} = -1 \)
- \( \begin{vmatrix} 2 & 1 \\ 0 & -1 \end{vmatrix} = -2 \)
- \( \begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} = 2 \)

Thus,
\[ \text{Del}_1 = 4(-1) + 2 + 3(2) = -4 + 2 + 6 = 4 \neq 0 \]

- **Calculating \( \text{Del}_2 \)**:
\[ \text{Del}_2 = \begin{vmatrix} 1 & 4 & 3 \\ 1 & 2 & 1 \\ 1 & 0 & -1 \end{vmatrix} \]
Following similar steps, we find that \( \text{Del}_2 \) also results in a non-zero value.

5. **Conclusion:**
Since \( \text{Del} = 0 \) and \( \text{Del}_1 \neq 0 \), we conclude that the system has infinitely many solutions.