Question 1105260: Solve the system of equations using matrices. Use Gaussian elimination with back-substitution. x + y + z = -5
x - y + 3z = -1
4x + y + z = -2
Found 2 solutions by greenestamps, richwmiller:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
I'm not sure what you mean by "Gaussian elimination with back substitution". Gaussian elimination solves the system; there is no substitution required.
You have the required "1" in position (1,1). Use it to get 0 in positions (2,1) and (3,1):
R2 <-- R1-R2; R3 <-- R3-4*R1: 
Take out the common factors in rows 2 and 3: 
Use the 1 in position (2,2) to get 0 in positions (1,2) and (3,2):
R1 <-- R1-R2; R3 <-- R3-R2: 
Take out the common factor in row 3: 
Use the 1 in position (3,3) to get 0 in positions (1,3) and (2,3):
R1 <-- R1-2*R3; R2 <-- R2+R3: 
We are done; the system has a unique solution
x = 1; y = -4, z = -2
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! back substitution would be to take the solution in row three and apply it to row two
then take the solution in row two and apply it to row one
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