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CSC 412 HW1 (Sample) Marcus Schaefer Problem 2 The system to solve is x -2 y = 5 -2x+4y = -10 Applying E 21 (2) yields: x – 2y = 5 0 = 0 So the system is consistent, and the solution is not unique. All solutions can be found by choosing y arbitrarily, and then letting x = 5+2y. Here is verification of my work using Matlab: Or here is another way of presenting the solution: Problem 2
Transcript of ovid.cs.depaul.eduovid.cs.depaul.edu/Classes/CSC412-W20/hwexample.d… · Web viewCSC 412 HW1...
![Page 1: ovid.cs.depaul.eduovid.cs.depaul.edu/Classes/CSC412-W20/hwexample.d… · Web viewCSC 412 HW1 (Sample) Marcus Schaefer Problem 2 The system to solve is x -2 y = 5-2x+4y = -10 Applying](https://reader036.fdocuments.us/reader036/viewer/2022070807/5f05c2e87e708231d414933f/html5/thumbnails/1.jpg)
CSC 412HW1 (Sample)
Marcus Schaefer
Problem 2
The system to solve is
x -2 y = 5 -2x+4y = -10
Applying E21(2) yields:
x – 2y = 5 0 = 0
So the system is consistent, and the solution is not unique. All solutions can be found by choosing y arbitrarily, and then letting x = 5+2y. Here is verification of my work using Matlab:
Or here is another way of presenting the solution:
Problem 2
I did the problem by hand (with Matlab verification):
![Page 2: ovid.cs.depaul.eduovid.cs.depaul.edu/Classes/CSC412-W20/hwexample.d… · Web viewCSC 412 HW1 (Sample) Marcus Schaefer Problem 2 The system to solve is x -2 y = 5-2x+4y = -10 Applying](https://reader036.fdocuments.us/reader036/viewer/2022070807/5f05c2e87e708231d414933f/html5/thumbnails/2.jpg)
Problem 3
3a: The augmented matrix is as follows:
3b) Applying E4(1/2) yields:
Next step is E34(1):
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and so on ….