Question 152209:  solve the system using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination: 
 
w+x+y+z=5 
w+2x-y-2z=-1 
w-3x-3y-z=-1 
2w-x+2y-z=-2 
 Found 2 solutions by  nabla, richwmiller: Answer by nabla(475)      (Show Source): 
You can  put this solution on YOUR website! 1 1   1  1  5 
1 2  -1 -2 -1 
1 -3 -3 -1 -1 
2 -1  2 -1 -2
 
-R1+R2, -R1+R3 
1 1   1  1  5 
0 1  -2 -3 -6 
0 -4 -4 -2 -6 
2 -1  2 -1 -2
 
-2R1+R4 
1 1   1  1  5 
0 1  -2 -3 -6 
0 -4 -4 -2 -6 
0 -3  0 -3 -12
 
-1/3R4, -1/2R3 
1  1  1  1  5 
0  1 -2 -3 -6 
0  2  2  1  3 
0  1  0  1  4
 
-R2+R1, -R2+R4 
1  0  3  4  11 
0  1 -2 -3 -6 
0  2  2  1  3 
0  0  2  4  10
 
-2R2+R3 
1  0  3  4  11 
0  1 -2 -3 -6 
0  0  6  7  15 
0  0  2  4  10
 
1/2R4 <-> R3 
1  0  3  4  11 
0  1 -2 -3 -6 
0  0  1  2  5 
0  0  6  7  15
 
-3R3+R1,2R3+R2, -6R3+R4 
1  0  0 -2  -4 
0  1  0  1   4 
0  0  1  2   5 
0  0  0 -5  -3
 
-1/5R5 
1  0  0 -2  -4 
0  1  0  1   4 
0  0  1  2   5 
0  0  0  1  3/5
 
-2R4+R3,-R4+R2, 2R4+R1 
1  0  0  0 -14/5 
0  1  0  0  17/5 
0  0  1  0  19/5 
0  0  0  1  3/5
 
 
This gives solution set {-14/5, 17/5, 19/5, 3/5}. 
 Answer by richwmiller(17219)      (Show Source): 
You can  put this solution on YOUR website! 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 
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