Solve the following system of equations using Gaussian Elimination Method

Question

Solve the following system of equations using Gaussian Elimination Method $$x+2y+3z=2$$ $$x+y-z=1$$ $$2x+3y+2z=3$$.

My Attempt : $$x+2y+3z=2………(1)$$ $$x+y-z=1…………(2)$$ $$2x+3y+2z=3………(3)$$ Subtracting equation $(1)$ from equation $(2)$, we have $$y+4z=1………(4)$$ Multiplying equation $(1)$ by $2$ and then Subtracting from equation $(3)$, we have $$y+4z=1………(5)$$ Subtracting equation $(4)$ from equation $(5)$, we have $$0=0$$.

How do I proceed now?

Answer 1

You will get the system: $$x+2y+3z=2$$ $$-y-4z=-1$$ Substituting $$z=t$$ you will get $$y=1-4t$$ and $$x=2-2(1-4t)-3t$$ The system has infinity many solutions.

Answer 2

Since adding first two you are getting last one the system has infinite solutions.

From $ y=1-4z$ you get, by puting it in (2) $x =5z$. So your system has a solution $$(x,y,z) = (5t,1-4t,t)$$ where $t$ is an arbitray real number.