- For 1a solve with back substitution!
- For 3 you need to create a system that has this solution! There is not one answer to this--there are many, many answers! I would recommend making up a system that is in row echelon form--so start with z = 0 (bottom equation). Then, write an equation that incorporates y and z so that y = 1 (middle equation). Finally, write a (top) equation that incorporates x, y, and z so that x = 2 (given that y = 1 and z = 0).
- It is also possible to answer this with a system that is not in row echelon form.
- If you write two answers to question c--one system with the solution (2,1,0) in row echelon form, and another system with this solution that has an x,y, and z term for each equation--you will receive homework extra credit! You will have to turn this in to me tomorrow!
- For 5, even though it looks complex (because it's an application problem), your challenge is to solve that given system using your row operations! Good luck!
- If you were out today we worked on questions 2a and 2b in class--do those and you're all caught up!
Here are the answers so you can check yours:
1.) (-2, -3.3333, -4)
2a.) (-2, -3, 3)
2b.) (-5, -5, 3)
3.) Answer will vary (we'll discuss tomorrow)
5.) (1, 2, 1)
See you all tomorrow for some new stuff!
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