The system of equations `x + y + z=2, 3x – y +2z=6 and 3x + y +z=-18` has

Options

A. a unique solution
B. no solution
C. infinite no. of solutions
D. none of these

Answer

To determine the nature of the solutions for the given system of equations, we can use the determinant method. We will set up the equations and form the coefficient matrix, then calculate the determinant.

**Given equations:**
1. \( x + y + z = 2 \) (Equation 1)
2. \( 3x - y + 2z = 6 \) (Equation 2)
3. \( 3x + y + z = -18 \) (Equation 3)

**Step 1: Write the system in matrix form.**

The system can be represented in the form \( AX = B \), where:
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & -1 & 2 \\ 3 & 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 2 \\ 6 \\ -18 \end{bmatrix} \]

**Step 2: Calculate the determinant of matrix A.**

To find the determinant of matrix \( A \):
\[ \text{det}(A) = \begin{vmatrix} 1 & 1 & 1 \\ 3 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} \]

Using the rule of Sarrus or cofactor expansion, we can calculate the determinant:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 2 \\ 3 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & -1 \\ 3 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} = (-1)(1) - (2)(1) = -1 - 2 = -3 \)
2. \( \begin{vmatrix} 3 & 2 \\ 3 & 1 \end{vmatrix} = (3)(1) - (2)(3) = 3 - 6 = -3 \)
3. \( \begin{vmatrix} 3 & -1 \\ 3 & 1 \end{vmatrix} = (3)(1) - (-1)(3) = 3 + 3 = 6 \)

Now substituting back into the determinant:
\[ \text{det}(A) = 1(-3) - 1(-3) + 1(6) = -3 + 3 + 6 = 6 \]

**Step 3: Analyze the determinant.**

Since \( \text{det}(A) = 6 \neq 0 \), the system of equations has a unique solution.

**Conclusion:**
The system of equations has a unique solution.

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