Solving via Gaussian Elimination

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mad_monkey_j
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Homework Statement


Solve the following system of equations using Gaussian Elimination:
-2a+5c=1
a+2b-c=2
3a-2b=3

Homework Equations


N/A

The Attempt at a Solution


Augmented matrix
(-2 0 5 | 1) R2 → R2 +R1/2
(1 2 -1 | 2) R3 → R3 - 3R2
(3 -2 0 | 3)

(-2 0 5 | 1) Swap R2 and R3
(0 2 3/2 | 5/2)
(0 -8 0 | -3)

(-2 0 5 | 1) R2 → -R2/4
(0 -8 0 | -3)
(0 2 3/2 | 5/2)

(-2 0 5 | 1) R3 → R3-R2
(0 2 0 | 3/4)
(0 2 3/2 | 5/2)

(-2 0 5 | 1) R3 → R3*10/3
(0 2 0 | 3/4)
(0 2 3/2 | 5/2)

(-2 0 5 | 1) R1 → R1-R3
(0 2 0 | 3/4)
(0 0 5 | 7/12)

(-2 0 0 | 5/12)
(0 2 0 | 3/4)
(0 0 5 | 7/12)

therefore a=-5/24, b=3/8 and c=7/60

The problem is a and c satisfy the equation but b does not, not sure what I've done wrong especially since a and c both work. Any help?
 
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verty said:
You may not do two operations simultaneously.
?? He didn't.

mad_monkey_j said:

Homework Statement


Solve the following system of equations using Gaussian Elimination:
-2a+5c=1
a+2b-c=2
3a-2b=3

Homework Equations


N/A

The Attempt at a Solution


Augmented matrix
(-2 0 5 | 1) R2 → R2 +R1/2
(1 2 -1 | 2) R3 → R3 - 3R2
(3 -2 0 | 3)
Your error is here. R3- 3R2, in the third column, would be 0-3(-1)= 3, not 0.

(-2 0 5 | 1) Swap R2 and R3
(0 2 3/2 | 5/2)
(0 -8 0 | -3)

(-2 0 5 | 1) R2 → -R2/4
(0 -8 0 | -3)
(0 2 3/2 | 5/2)

(-2 0 5 | 1) R3 → R3-R2
(0 2 0 | 3/4)
(0 2 3/2 | 5/2)

(-2 0 5 | 1) R3 → R3*10/3
(0 2 0 | 3/4)
(0 2 3/2 | 5/2)

(-2 0 5 | 1) R1 → R1-R3
(0 2 0 | 3/4)
(0 0 5 | 7/12)

(-2 0 0 | 5/12)
(0 2 0 | 3/4)
(0 0 5 | 7/12)

therefore a=-5/24, b=3/8 and c=7/60

The problem is a and c satisfy the equation but b does not, not sure what I've done wrong especially since a and c both work. Any help?
 
HallsofIvy said:
?? He didn't.


Your error is here. R3- 3R2, in the third column, would be 0-3(-1)= 3, not 0.

Ahh thanks can't believe I didn't notice it, all works nicely now.