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anonymous

  • one year ago

Solve the system. x1 - 5x2 + 4x3 = -3 2x1 - 7x2 + 3x3 = -2 -2x1 + x2 + 7x3 = -1

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  1. anonymous
    • one year ago
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    I know that we are to end up with a matrices in the form of 1 ? ? ? 0 1 ? ? 0 0 1 ?

  2. Hero
    • one year ago
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    @LadyInkblot have you converted the system to a matrix yet?

  3. anonymous
    • one year ago
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    Yes. It starts out as a matrix of 1 -5 4 -3 2 -7 3 -2 -2 1 7 -1

  4. Hero
    • one year ago
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    How far have you gotten with reducing the matrix?

  5. anonymous
    • one year ago
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    I don't know if this is correct, but so far I have 1 -5 4 -3 0 3 -5 4 0 -10 15 -7

  6. Hero
    • one year ago
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    Looks like you attempted to perform a couple of row operations. Do you remember which row operations you performed?

  7. anonymous
    • one year ago
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    I preformed (I believe) 2R1 + R3 and -2R1 - R2

  8. Hero
    • one year ago
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    I can double check that for you. In the mean time, do you know how you plan to continue reducing the matrix from here?

  9. anonymous
    • one year ago
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    I do not know how to proceed, which is why I came here for help ^_^;

  10. Hero
    • one year ago
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    May I suggest a different approach?

  11. anonymous
    • one year ago
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    I know that I need to get rid of the -10x2, but I'm not sure how to go about it, since I don't want anything to go back into my third row first column spot.

  12. anonymous
    • one year ago
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    Of course!

  13. Hero
    • one year ago
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    What if, starting with the original matrix, you performed the following operations to start with: R2 + R3, then 2R1 + R2

  14. anonymous
    • one year ago
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    Just row 2 + row 3? not multiplying anything?

  15. Hero
    • one year ago
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    Yes, actually do R2 + R3, then -2R1 + R2

  16. anonymous
    • one year ago
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    okay. I'll get back to you in a minute then with my answer...

  17. anonymous
    • one year ago
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    Okay, so I got 1 -5 4 -3 0 3 -5 0 0 6 -4 -4

  18. anonymous
    • one year ago
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    But then I'll still have to get rid of the 6 in the third row... so what about 2R2 + r3?

  19. Hero
    • one year ago
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    You should double check your work for those 2 operations.

  20. anonymous
    • one year ago
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    okay, what's wrong with it?

  21. anonymous
    • one year ago
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    Okay, I tried again and got something different 1 -5 4 -3 0 3 -5 -8 0 -6 10 -3

  22. Hero
    • one year ago
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    The 1st and 3rd row are correct. The second row has a mistake with the -8

  23. anonymous
    • one year ago
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    Ah. Would it be a 4?

  24. Hero
    • one year ago
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    Yes

  25. anonymous
    • one year ago
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    1 -5 4 -3 0 3 -5 4 0 0 0 5

  26. Hero
    • one year ago
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    You're jumping too far ahead.

  27. Hero
    • one year ago
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    Please post just the first two operations.

  28. anonymous
    • one year ago
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    1 -5 4 -3 0 3 -5 4 0 -6 10 -3

  29. Hero
    • one year ago
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    Okay, yes, and in the next step, you figure out that you have a row of the form 0 0 0 b which is an invalid row.

  30. anonymous
    • one year ago
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    Yes. What does that mean, an invalid row. 1 -5 4 -3 0 3 -5 4 0 0 0 5

  31. Hero
    • one year ago
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    If you converted the last row back to an algebraic equation, you'd have the form 0x + 0y + 0z = 5 or just 0 = 5, but 0 = 5 is a false statement.

  32. anonymous
    • one year ago
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    Okay. So what does that mean for the answer to the problem then. Just that the solution is invalid?

  33. Hero
    • one year ago
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    It means the system cannot be solved and therefore has no solution.

  34. anonymous
    • one year ago
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    Okay, that makes sense. Thank you!

  35. Hero
    • one year ago
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    You're welcome. Great work on your problem.

  36. anonymous
    • one year ago
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    Thank you!

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