Question:
The system of linear equations \(2x+y+z=0,\ y-z=0,\ x+y=0\) has
The system of linear equations \(2x+y+z=0,\ y-z=0,\ x+y=0\) has
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If a homogeneous system has a free parameter, then it has infinitely many solutions.
Updated On: Jul 3, 2026
- An infinite number of solutions
- No solution
- A unique solution
- Two solutions
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The Correct Option is A
Solution and Explanation
Concept:
A homogeneous system of linear equations always has at least the trivial solution. If the equations are dependent, then the system has infinitely many solutions.
Step 1: Write the equations. \[ 2x+y+z=0 \] \[ y-z=0 \] \[ x+y=0 \]
Step 2: Solve using equations. From \[ y-z=0 \] we get \[ y=z \] From \[ x+y=0 \] we get \[ x=-y \] Substitute in first equation: \[ 2(-y)+y+y=0 \] \[ -2y+y+y=0 \] \[ 0=0 \] So the first equation is dependent on the other two.
Step 3: Express solution. Let \[ y=t \] Then, \[ x=-t,\qquad z=t \] So solutions are \[ (x,y,z)=(-t,t,t) \] where \(t\in R\). Thus there are infinitely many solutions.
Step 4: Final answer. \[ \boxed{\text{An infinite number of solutions}} \]
A homogeneous system of linear equations always has at least the trivial solution. If the equations are dependent, then the system has infinitely many solutions.
Step 1: Write the equations. \[ 2x+y+z=0 \] \[ y-z=0 \] \[ x+y=0 \]
Step 2: Solve using equations. From \[ y-z=0 \] we get \[ y=z \] From \[ x+y=0 \] we get \[ x=-y \] Substitute in first equation: \[ 2(-y)+y+y=0 \] \[ -2y+y+y=0 \] \[ 0=0 \] So the first equation is dependent on the other two.
Step 3: Express solution. Let \[ y=t \] Then, \[ x=-t,\qquad z=t \] So solutions are \[ (x,y,z)=(-t,t,t) \] where \(t\in R\). Thus there are infinitely many solutions.
Step 4: Final answer. \[ \boxed{\text{An infinite number of solutions}} \]
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