Solving via Gaussian Elimination

In summary, the system of equations can be solved using Gaussian Elimination. However, care must be taken to avoid doing two operations simultaneously. The final solution is a = -5/24, b = 3/8, and c = 7/60.
  • #1
mad_monkey_j
33
0

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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  • #2
-2a+5c=1
-2(-5/24) + 5(7/60) = 10/24+30/60 = 5/12+6/12

so a & c do not work, I would go back and check each of your steps
 
  • #3
You may not do two operations simultaneously.
 
  • #4
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?
 
  • #5
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.
 

Related to Solving via Gaussian Elimination

1. What is Gaussian Elimination?

Gaussian Elimination is a method used to solve systems of linear equations. It involves using row operations to transform the equations into simpler forms until a solution can be easily obtained.

2. How does Gaussian Elimination work?

Gaussian Elimination works by using three types of row operations: multiplication by a constant, adding one row to another, and swapping rows. These operations are performed in a specific order to transform the equations into a triangular form, making it easier to solve for the variables.

3. What are the advantages of using Gaussian Elimination?

Gaussian Elimination is a systematic and efficient method for solving systems of equations. It eliminates the need for guesswork and reduces the chances of errors. It also allows for the solution to be easily verified and can be used to solve systems with any number of variables.

4. Are there any limitations of Gaussian Elimination?

One limitation of Gaussian Elimination is that it can become numerically unstable when dealing with large systems of equations or when working with equations that have very small or very large coefficients. It also cannot be used to solve systems of equations with no solutions or infinitely many solutions.

5. How is Gaussian Elimination used in real-world applications?

Gaussian Elimination is commonly used in engineering, physics, and other scientific fields to solve systems of equations that represent real-world problems. It can be used to analyze circuits, model chemical reactions, and solve optimization problems. It is also used in computer graphics to solve linear equations for 3D animation and game development.

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