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Imtiaz7

  • one year ago

Help Please...!!!

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  1. Imtiaz7
    • one year ago
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    Convert following matrix into Reduced Echlon Form:

  2. Imtiaz7
    • one year ago
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    |dw:1438863140426:dw|

  3. UsukiDoll
    • one year ago
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    just reduce echelon form right?

  4. UsukiDoll
    • one year ago
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    so what we need is an upper triangular matrix which means that below the main diagonal we need all 0's we can achieve this by using row operations by swapping rows, multiplying a row by a scalar number, or add/subtract rows and replace the old row with the new result after that computation

  5. UsukiDoll
    • one year ago
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    keep in mind that we need a 1 always at the top left the 1 also has to go towards the right and the bottom... we also need a row of zeros

  6. Imtiaz7
    • one year ago
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    wait I've solve this, can you please tell me is this right?

  7. UsukiDoll
    • one year ago
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    |dw:1438863440670:dw| so the places I've circled is the main diagonal so we need to get rid of the 3 , -1, and 4 using row operations... yeah sure.

  8. UsukiDoll
    • one year ago
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    anyway... -3r1 + r2 -> r2 r1+ r3 - > r3

  9. UsukiDoll
    • one year ago
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    |dw:1438863691395:dw||dw:1438863716259:dw|

  10. UsukiDoll
    • one year ago
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    |dw:1438863747952:dw|

  11. UsukiDoll
    • one year ago
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    hmmm gonna make life a bit easier by 1/2r3->r3

  12. UsukiDoll
    • one year ago
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    |dw:1438863802231:dw|

  13. UsukiDoll
    • one year ago
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    -r2+11r3 ->r3

  14. UsukiDoll
    • one year ago
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    |dw:1438863836507:dw|

  15. UsukiDoll
    • one year ago
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    1/10 r3 - >r3 1/11 r2 - > r2

  16. UsukiDoll
    • one year ago
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    |dw:1438863882447:dw|

  17. UsukiDoll
    • one year ago
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    do you have a matrix that looks like this? ^

  18. Imtiaz7
    • one year ago
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  19. UsukiDoll
    • one year ago
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    how did you get 20/11 ?

  20. Imtiaz7
    • one year ago
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    by multiplying Row 2 by -2 and add it in row 3(-2*1/11 + 2)

  21. UsukiDoll
    • one year ago
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    hmmm dealing with fractions that early is killer though

  22. UsukiDoll
    • one year ago
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    because what I've done to the third row was divide by 1/2 all over row 3 so I would have just 0 1 1

  23. UsukiDoll
    • one year ago
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    so when I had 0 11 1 0 1 1 I distributed the negative sign on row 2 and then multiply row 3 with 11 -r2+11r^3 -> r^3

  24. Imtiaz7
    • one year ago
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    Uhm so i i did wrong?

  25. UsukiDoll
    • one year ago
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    0 -11 -1 0 11 11 adding them together would make it 0 0 10 and then 1/10r3 ->r3 to get rid of that 10 to have 0 0 1

  26. Imtiaz7
    • one year ago
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    but in last it becomes in reduced echlon form; isnt?

  27. UsukiDoll
    • one year ago
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    and then to have the 1 in the middle I had to divide the whole second row by 1/11

  28. UsukiDoll
    • one year ago
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    somehow our first rows are different but the second and third is the same..

  29. Imtiaz7
    • one year ago
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    does my answer is correct?

  30. UsukiDoll
    • one year ago
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    it should be as long as it followed the reduce echelon rules.. and it does. we do have an upper triangular matrix and our 1 is going down and shifted to the right

  31. Imtiaz7
    • one year ago
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    Yes! So our answers are different, but rules are same!

  32. UsukiDoll
    • one year ago
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    hmm I'm assuming that an extra step was taken to get rid of the -2 on the first row. unless we are required to get RREF form, that's not required. What's required is the numbers until the main diagonal go to 0.

  33. UsukiDoll
    • one year ago
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    check this one out... it also did something different but ended up with the same thing before taking RREF form oh that 1 in the third row got blocked out, but noticed how this calculator did something I disagree with, but still ends up following the rules?

  34. Imtiaz7
    • one year ago
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    wait let me check it

  35. UsukiDoll
    • one year ago
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    that was done by a calculator bot... I really hated that 0 0 54/11 but by multiplying 11/54r3 -> r3 0 0 1

  36. Imtiaz7
    • one year ago
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    @UsukiDoll the question is different there, that is there is 1 instead of 1 in first row

  37. Imtiaz7
    • one year ago
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    -1*

  38. UsukiDoll
    • one year ago
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    ugh a typo.. yeah this is way late at night... on my end.

  39. UsukiDoll
    • one year ago
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    one sec.

  40. UsukiDoll
    • one year ago
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  41. UsukiDoll
    • one year ago
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    still has that big fraction 20/11 on row 3 but 11/20r3 - >r3 makes it 0 0 1

  42. Imtiaz7
    • one year ago
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    my answer is correct!!!

  43. UsukiDoll
    • one year ago
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    mhm

  44. Imtiaz7
    • one year ago
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    :) lol Thank you so much @UsukiDoll to help me..

  45. UsukiDoll
    • one year ago
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    XD you're welcome... these questions always come up very late at night, so typos are bound to happen... typed too fast XD

  46. Imtiaz7
    • one year ago
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    Its okay, when i post i dont run xD

  47. UsukiDoll
    • one year ago
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    yay I got it right too XD |dw:1438865533703:dw| row operation #5 on the correct image has the same matrix I had... only in fewer steps.

  48. UsukiDoll
    • one year ago
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    but as long as the rules aren't broken for row echelon form, then everything will be fine ^_^

  49. Imtiaz7
    • one year ago
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    Yup @UsukiDoll rules are for obey uhm i think so lol

  50. Imtiaz7
    • one year ago
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    @UsukiDoll ?

  51. UsukiDoll
    • one year ago
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    hmmm?

  52. Imtiaz7
    • one year ago
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    There is still a confusion uhm the answer you showed me are correct one but in my answer in fist row there is -9/11, still is it right?

  53. UsukiDoll
    • one year ago
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    you mean for this right http://assets.openstudy.com/updates/attachments/55c34eede4b06d80932e63a6-imtiaz7-1438864107356-20150806_172006.jpg ?

  54. UsukiDoll
    • one year ago
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    as long as you have an upper triangular matrix which means all 0's below the main diagonal and have a 1 on the top left that goes to the bottom and right... I don't see a problem.

  55. Imtiaz7
    • one year ago
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    Okay :)

  56. Imtiaz7
    • one year ago
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    Once again thanks a lot :)

  57. UsukiDoll
    • one year ago
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    we could RREF that last matrix with that -9/11 if you want

  58. UsukiDoll
    • one year ago
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    to show that we can get an identity matrix out of some more row operations..

  59. UsukiDoll
    • one year ago
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    |dw:1438866425233:dw| this was the matrix you had in REF form right?

  60. Imtiaz7
    • one year ago
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    Yes

  61. UsukiDoll
    • one year ago
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    I would use row 3 to get rid of the remaining numbers above the main diagonal, but I hate dealing with fractions so 11r2 - >r2 11r1 -> r1

  62. UsukiDoll
    • one year ago
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    |dw:1438866522476:dw|

  63. UsukiDoll
    • one year ago
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    -r3+r2 - > r2 -9r3 + r1 - > r1

  64. UsukiDoll
    • one year ago
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    |dw:1438866568376:dw||dw:1438866589363:dw|

  65. Imtiaz7
    • one year ago
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    @UsukiDoll can i give you another medal bc your math is too good

  66. UsukiDoll
    • one year ago
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    1/11r2 - >r2 1/11 r1 -> r1... I wish we can give more than 1 medal XD

  67. UsukiDoll
    • one year ago
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    |dw:1438866634762:dw| a-ha identity matrix! so we got RREF XD

  68. Imtiaz7
    • one year ago
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    lol

  69. UsukiDoll
    • one year ago
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    which means your row operations from earlier were correct... otherwise we wouldn't get RREF

  70. Imtiaz7
    • one year ago
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    yes...

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