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anonymous

  • one year ago

Can someone look at this and tell me what I did wrong to get this answer??? Will medal!

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    This took me so long to do I'm so upset... I was so proud of myself until I got that answer :(

  3. UsukiDoll
    • one year ago
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    why does the formula require the inverse? Wait I see what the problem is. You don't have to include x y z you just include the numbers that are attached to the x y z in the set of linear equation s

  4. anonymous
    • one year ago
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    um, explain?

  5. anonymous
    • one year ago
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    someone told me that the formula was A^-1*B...

  6. anonymous
    • one year ago
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    and when I looked it up that was correct, using the inverse

  7. amistre64
    • one year ago
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    inverses are such a pain, when they get big its sometimes more efficient to use augmented form.

  8. UsukiDoll
    • one year ago
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    I think something happened to your calculations.. your x is huge

  9. UsukiDoll
    • one year ago
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    that's what I was trying to say... put it in an augmented matrix format and use row operations...

  10. anonymous
    • one year ago
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    I don't remember learning that I don't understand!

  11. amistre64
    • one year ago
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    rref{ {1,1,1,4}, {4,5,0,3}, {0,1,-3,-10} } http://www.wolframalpha.com/input/?i=rref%7B+%7B1%2C1%2C1%2C4%7D%2C+%7B4%2C5%2C0%2C3%7D%2C+%7B0%2C1%2C-3%2C-10%7D+%7D

  12. amistre64
    • one year ago
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    less room for error in my opinion

  13. UsukiDoll
    • one year ago
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    do you see the original system of equations? YOu just put the numbers from the systems of equations like this 1 1 1 l 4 4 5 0 l 3 0 1 -3 l -10 and use row operations... you don't have a 0 in the main diagonal so row swapping isn't required.

  14. anonymous
    • one year ago
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    just tell me what I did wrong so I can fix it and move on! I've been working on this problem for HOURS

  15. anonymous
    • one year ago
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    lol math making people crazy when ever they can't find the answer ;D

  16. anonymous
    • one year ago
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    I don't understand :'(

  17. anonymous
    • one year ago
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    nothing nothing they are going help :D

  18. amistre64
    • one year ago
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    your inverse has one bad spot ... row1 col3: should be -5

  19. anonymous
    • one year ago
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    I used a calculator to find the inverse.... it shouldn't be wrong..... but I'll double check

  20. UsukiDoll
    • one year ago
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    I would've used augmented version and use row operations..

  21. amistre64
    • one year ago
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    it prolly had a negative, and you might have missed it ... but thats the trouble spot

  22. anonymous
    • one year ago
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    oh... you're right @amistre64 I wrote it wrong...

  23. UsukiDoll
    • one year ago
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    I've made plenty of mistakes on my exams xD! It's not a problem. xD

  24. anonymous
    • one year ago
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    thank you so much... @amistre64 I'll calculate it now and look at my answer to see if its correct this time

  25. amistre64
    • one year ago
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    -50 instead of +50, so your 100 off on the first row :) 100-98 = 2

  26. amistre64
    • one year ago
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    good effort

  27. anonymous
    • one year ago
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    haha thanks I'm not sure who to give the medal to... :P

  28. anonymous
    • one year ago
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    @amistre64 can you give @UsukiDoll a medal for me since I medaled you?

  29. UsukiDoll
    • one year ago
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    interesting.. so there is more than one method to solve this... my professor didn't even teach the inverse version... just told me to put it in augmented matrix form and row operations solve away until I have reduce row echelon form. So the final matrix wouldv'e looked like this 1 0 0 l 2 0 1 0 l-1 0 0 1 l 3 for x =2, y = -1, z=3

  30. anonymous
    • one year ago
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    Much better :D

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  31. UsukiDoll
    • one year ago
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    yay :)

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spraguer (Moderator)
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