Linear Systems in 3 Variables

(Consistent Independent System) 
x-2y+3z=9
-x+3y    =-4
2x-5y+5z=17

First Step:
*Add equation 1 and 2. That yields to, y+3z=5. 

Second Step:
*Mutiply (-2) to equation 1. -2(x-2y+3z), that yields to, 
(-2x+4y-6z=-18).

Third Step:
*Add the result in step 2 to equation number 3 to cancel the leading coefficient.
    -2x+4y-6z=-18
 + 2x-5y+5z=17 , this results to (-y-z=-1).

Fourth Step:
*Add the result in step 2 to the result in step 1.
    -y-z=-1
+ y+3z=5 , this results to (2z=4) , (z=2).

Fifth Step: 
*Substitute the values into the equation order to get the Solution Set of the system. 
Solution: 
x-2(-1)+3(2)=9
          x+2+6=9
                   x=1 
Solution Set:
  (1,-1,2)  *There is only one solution. 


Row Echelon Form: 
x-2y+3z=9
     y+3z=5
           z=2

(Inconsistent System) 
x1-3x2+x3=1
2x1-x2-2x3=2
x1+2x2-3x3=-1

First Step: 
*Multiply (-2) to equation 1 then add to equation 2. 
  -2x1+6x2-2x3=-2
+ 2x1-x2-2x3=2
*this results to, 5x2-4x3=0. 

Second Step:
*Multiply (-2) to equation 3 then add to equation 2. 
     -2x1+4x2-6x3=2
  + 2x1-x2-2x3=2
*this results to, -5x2+4x3=4. 

Third Step: 
*Add the results in the first and second step. 
    5x2-4x3=0
+-5x2+4x3=4 
*this results to, 0=4. (There is no solution in this system) 

Row Echelon Form: 
x1-3x2+x3=1
     5x2-4x3=0
              0=4 

(Consistent Dependent) 
      x2-x3=0
x1       -x3=-1
-x1+3x2   =1

First Step: 
*Interchange equations 1 and 2. 
x1       -x3=-1
       x2-x3=0
-x1+3x2   =1

Second Step: 
*Multiply (-3) to equation 2. 
   3x2+3x3=0

Third Step:
*Add equations 1 and 2. 
   3x2-3x3=0
+ 3x2+3x3=0
              0=0
Fourth Step: 
*Rewrite in a system of linear equation. 
x1-3x3=-1
    3x2-3x3=0
              0=0


Parametric Representation: 
x1-3x3=-1
x1=-1+3x3 free variable   
Let x3=f, f is a real number. 
x1=3f-1
x2=f
x3=f, f=2

The solution set is: (5,2,2) or (2,1,1) 
  Let