Question

The system of equations
\[ x+y+z=0 \] \[ 2x+3y+z=0 \] \[ x+2y=0 \] has

• A. a unique solution; \(x=0,y=0,z=0\)
• B. infinite solutions
• C. no solution
• D. finite number of non-zero solutions

Hint:

If one equation becomes an identity after substitution, system is dependent and gives infinitely many solutions.

Answer 1

The Correct Option is B

Step 1: Solve the third equation first.
\[ x+2y=0 \Rightarrow x=-2y \]
Step 2: Substitute \(x=-2y\) in the first equation.
\[ (-2y)+y+z=0 \Rightarrow -y+z=0 \Rightarrow z=y \]
Step 3: Substitute \(x=-2y\) and \(z=y\) in the second equation.
\[ 2(-2y)+3y+y=0 \Rightarrow -4y+4y=0 \Rightarrow 0=0 \]
So second equation is dependent.
Step 4: Final form of solution.
Let \(y=t\). Then:
\[ x=-2t,\quad y=t,\quad z=t \]
So infinitely many solutions exist.
Final Answer:
\[ \boxed{\text{Infinite solutions}} \]