Question 34772: Solve the following system of equations using Gaussian elimination (with a matrix) and back substitution, or by using Gauss-Jordan elimination.
3x + y - z = 0
x + y = 2z = 6
2x + 2y = 3z = 10
Found 2 solutions by checkley71, richwmiller:
Answer by checkley71(8403)   (Show Source):
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! if you mean
3,1,-1,0
1,1,+2,6
2,2,+3,10
divide row 1 by 3/1
1,1/3,-1/3,0
1,1,2,6
2,2,3,10
add down (-1/1) *row 1 to row 2
1,1/3,-1/3,0
0,2/3,7/3,6
2,2,3,10
add down (-2/1) *row 1 to row 3
1,1/3,-1/3,0
0,2/3,7/3,6
0,4/3,11/3,10
divide row 2 by 2/3
1,1/3,-1/3,0
0,1,21/6,9
0,4/3,11/3,10
add down (-4/3) *row 2 to row 3
1,1/3,-1/3,0
0,1,7/2,9
0,0,-1,-2
divide row 3 by -1/1
1,1/3,-1/3,0
0,1,7/2,9
0,0,1,2
add up (-7/2) *row 3 to row 2
1,1/3,-1/3,0
0,1,0,2
0,0,1,2
add up (1/3) *row 3 to row 1
1,3/9,0,2/3
0,1,0,2
0,0,1,2
add up (-1/3) *row 2 to row 1
1,0,0,0
0,1,0,2
0,0,1,2
final
1,0,0,0
0,1,0,2
0,0,1,2
1,0,0,0
0,1,0,2
0,0,1,2
if you mean
3,1,-1,0
1,1,-2,6
2,2,-3,10
divide row 1 by 3/1
1,1/3,-1/3,0
1,1,-2,6
2,2,-3,10
add down (-1/1) *row 1 to row 2
1,1/3,-1/3,0
0,2/3,-5/3,6
2,2,-3,10
add down (-2/1) *row 1 to row 3
1,1/3,-1/3,0
0,2/3,-5/3,6
0,4/3,-7/3,10
divide row 2 by 2/3
1,1/3,-1/3,0
0,1,-15/6,9
0,4/3,-7/3,10
add down (-4/3) *row 2 to row 3
1,1/3,-1/3,0
0,1,-5/2,9
0,0,1,-2
divide row 3 by 1/1
1,1/3,-1/3,0
0,1,-5/2,9
0,0,1,-2
add up (5/2) *row 3 to row 2
1,1/3,-1/3,0
0,1,0,4
0,0,1,-2
add up (1/3) *row 3 to row 1
1,3/9,0,-2/3
0,1,0,4
0,0,1,-2
add up (-1/3) *row 2 to row 1
1,0,0,-2
0,1,0,4
0,0,1,-2
final
1,0,0,-2
0,1,0,4
0,0,1,-2
1,0,0,-2
0,1,0,4
0,0,1,-2
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