Question 152209
w+x+y+z=5
w+2x-y-2z=-1
w-3x-3y-z=-1
2w-x+2y-z=-2

1,1,1,1,5
1,2,-1,-2,-1
1,-3,-3,-1,-1
2,-1,2,-1,-2

add  down (-1/1) *row 1 to row 2
1,1,1,1,5
0,1,-2,-3,-6
1,-3,-3,-1,-1
2,-1,2,-1,-2


add  down (-1/1) *row 1 to row 3
1,1,1,1,5
0,1,-2,-3,-6
0,-4,-4,-2,-6
2,-1,2,-1,-2



add  down (-2/1) *row 1 to row 4
1,1,1,1,5
0,1,-2,-3,-6
0,-4,-4,-2,-6
0,-3,0,-3,-12

add  down (4/1) *row 2 to row 3
1,1,1,1,5
0,1,-2,-3,-6
0,0,-12,-14,-30
0,-3,0,-3,-12


add  down (3/1) *row 2 to row 4
1,1,1,1,5
0,1,-2,-3,-6
0,0,-12,-14,-30
0,0,-6,-12,-30

divide row 3 by -12/1
1,1,1,1,5
0,1,-2,-3,-6
0,0,1,-14/-12,-30/-12
0,0,-6,-12,-30


add  down (6/1) *row 3 to row 4
1,1,1,1,5
0,1,-2,-3,-6
0,0,1,7/6,5/2
0,0,0,-5,-15

divide row 4 by -5/1
1,1,1,1,5
0,1,-2,-3,-6
0,0,1,7/6,5/2
0,0,0,1,3

This is where you would start back substitution.
We now have z=3
We continue with the matrix solution


add  up  (-7/6) *row 4 to row 3
1,1,1,1,5
0,1,-2,-3,-6
0,0,1,0,-1
0,0,0,1,3
now we have y= -1

add  up  (3/1) *row 4 to row 2
1,1,1,1,5
0,1,-2,0,3
0,0,1,0,-1
0,0,0,1,3


add  up  (-1/1) *row 4 to row 1
row 1 col 2
1,1,1,0,2
0,1,-2,0,3
0,0,1,0,-1
0,0,0,1,3


add  up  (2/1) *row 3 to row 2
1,1,1,0,2
0,1,0,0,1
0,0,1,0,-1
0,0,0,1,3
now we have  z=3 y=-1 x=1

add  up  (-1/1) *row 3 to row 1
1,1,0,0,3
0,1,0,0,1
0,0,1,0,-1
0,0,0,1,3

add  up  (-1/1) *row 2 to row 1
1,0,0,0,2
0,1,0,0,1
0,0,1,0,-1
0,0,0,1,3

now we have all four solutions
w=2 x=1 y=-1 z=3
(2,1,-1,3)

w+x+y+z=5
check
2+1-1-3=5
ok

w+2x-y-2z=-1
2+2-(-1)-6=-1
5-6=-1
ok
w-3x-3y-z=-1
2-3-3(-1)-3=-1
-1+3-3=-1
ok
2w-x+2y-z=-2
2*2-1+2(-1)-3=-2
4-1-2-3=-2
3-2-3=-2
ok

(2,1,-1,3) works in all 4 equations