Question:

The system of equation \(x - y + 3z = 0, x + z = 0, x + y - z = 0\) has _______

Show Hint

For any homogeneous system of equations, "no solution" is never a possible answer.
The system is always consistent because the trivial solution $(0,0,0)$ always satisfies it.
Updated On: Jul 9, 2026
  • a unique solution
  • finitely many solutions
  • infinitely many solutions
  • no solution
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
We need to determine the nature of the solution set for a system of three homogeneous linear equations in three variables. 

Step 2: Key Formula or Approach: 
A system of homogeneous linear equations can be written as: 
\[ AX = 0 \] A homogeneous system always has at least the trivial solution: \[ x=0,\quad y=0,\quad z=0 \] 
To check whether non-trivial solutions exist, we calculate the determinant of the coefficient matrix \(A\): 
- If \(\det(A) \neq 0\), the system has only the unique trivial solution. 
- If \(\det(A) = 0\), the system has infinitely many solutions. 

Step 3: Detailed Explanation: 

Step 3.1: Write the coefficient matrix \(A\): 
The coefficient matrix is: \[ A = \begin{pmatrix} 1 & -1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & -1 \end{pmatrix} \] 

Step 3.2: Calculate the determinant of \(A\): 
Expanding the determinant along the second row: \[ \det(A) = -1 \begin{vmatrix} -1 & 3 \\ 1 & -1 \end{vmatrix} + 0 - 1 \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \] 
Calculate the \(2 \times 2\) determinants: \[ \begin{vmatrix} -1 & 3 \\ 1 & -1 \end{vmatrix} = (-1)(-1)-(3)(1) \] \[ =1-3=-2 \] 
Similarly, \[ \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = (1)(1)-(-1)(1) \] \[ =1+1=2 \] 
Substituting these values: \[ \det(A) = -1(-2)-1(2) \] \[ =2-2 \] \[ \det(A)=0 \] 

Step 3.3: Analyze the Results: 
Since the determinant of the coefficient matrix is zero: \[ \det(A)=0 \] the system has non-trivial solutions. Therefore, the homogeneous system has infinitely many solutions. 

Step 4: Final Answer: 
The given system of homogeneous linear equations has: \[ \boxed{\text{Infinitely many solutions}} \]

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