Use the Gauss-Jordan elimination to solve the following system of equations?

Question

x+y+z=2
2x-3y+z=-11
-x+2y-z=8

Answer 1

[1 1 1 | 2]
[2 -3 1 | -11] -2*row 1
[-1 2 -1| 8] + row 1

[1 1 1 | 2]
[0 -5 -1 | -15] ÷ -5
[0 3 0 | 10]

[1 1 1 | 2] - row 2
[0 1 1/5 | 3]
[0 3 0 | 10] - 3*row 2

[1 0 4/5 | -1]
[0 1 1/5 | 3]
[0 0 -3/5 | 1] ÷ -3/5

[1 0 4/5 | -1] - 4/5*row 3
[0 1 1/5 | 3] - 1/5*row 3
[0 0 1 | -5/3]

[1 0 0 | 1/3]
[0 1 0 | 10/3]
[0 0 1 | -5/3]

x=1/3, y=10/3, z=-5/3

Answer 2

x+y+z=2 so
x=2-y-z so 2x = 4-2y-2z therefore
4-2y-2z -3y+z = -11 which we can simplify to
-5y-z= -15 or
5y+z = 15 so
z= 15-5y therefore
-(2-y-(15-5y)) + 2y - (15-5y) =8
-(2-15-y+5y) +2y -15 +5y = 8
-(-13+4y)+7y-15 =8
13-4y+7y-15=8
-2+3y=8
3y=10
y=10/3

going back to the first equation, we can solve for x
x + 10/3 +15-5(10/3)=2
x+10/3 +15-16-2/3=2
x+5/3=2
x=2-5/3
x=1/3

then solve for z
1/3 +10/3 +z = 2
z= 2-11/3
z=-5/3
___________
in summary:
x= 1/3
y= 10/3
z= -5/3
___________
to double check, plug them into the first equation
1/3 +10/3 -5/3 = 2
11/3-5/3=2
6/3= 2. . . yep!

Answer 3

The important thing is to understand how the equations are being solved. Performing matrix algebra in this forum is a bit awkward.