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jump Group Title

I need help solving this system of a linear equation based on the attached augmented MATRIX

  • one year ago
  • one year ago

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  1. jump Group Title
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    |dw:1355196230716:dw|

    • one year ago
  2. PhoenixFire Group Title
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    You need to reduce it to something like this: |dw:1355198633863:dw|

    • one year ago
  3. jump Group Title
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    yeh, i watched some video chalkboard examples, but they didn't have the bigger numbers in matrix! its got me vexed...lol

    • one year ago
  4. PhoenixFire Group Title
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    It's a thing called "Reduced Row Echelon Form". So firstly start by making --m(row)(column)-- m00 = 1 then m10 = 0 then m20=0 So the first colum will be 1, 0, 0

    • one year ago
  5. PhoenixFire Group Title
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    If you swap Row 2 with row 1 it'll put that 1 in the right place allowing you to make the rest of the column 0 easier.

    • one year ago
  6. jump Group Title
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    how do i get 100 with that starting matrix though 3,1,-2? canu show me the next couple steps on this problem?

    • one year ago
  7. PhoenixFire Group Title
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    I with this was easier to type on a computer =( |dw:1355199296801:dw|

    • one year ago
  8. jump Group Title
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    so i can just change the position of rowa 1 and 2?

    • one year ago
  9. PhoenixFire Group Title
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    Yes, but you have to change the ENTIRE row.. Hence why they're called "Row Operations" 1) Interchange two rows. 2) Multiple a row by any non-zero number 3) Add a multiple of a row to another row.

    • one year ago
  10. jump Group Title
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    yes, i find the draw pad fastest...lol

    • one year ago
  11. jump Group Title
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    i think i am mentally wounded on this. just been way too long since i had to do one!! hate to ask, but my answer choices are like (1,1,0). how do i get it rest of the way to that. i don't trust myself on this one if i did get to the end!@

    • one year ago
  12. jump Group Title
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    if its easier to work out on ur notepad,thats ok just pass me the 3 valuessoi can get out of this section. I'm definitely going to needto go back thru the tutorialfromthe start on these matrix'?

    • one year ago
  13. PhoenixFire Group Title
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    Only notebook I could find nearby, sorry it's hard to read but my values are: 1, -1, 0

    • one year ago
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  14. PhoenixFire Group Title
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    \[3x_1+x_2-2x_3=2\]\[3(1)+(-1)-2(0)=3-1=2\]

    • one year ago
  15. PhoenixFire Group Title
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    \[x_1-2x_2+x_3=3\]\[(1)-2(-1)+0=1+2=3\] \[2x_1-x_2-3x_3=3\]\[2(1)-(-1)-3(0)=2+1=3\] The values work in all equations therefore 1, -1, 0 is correct

    • one year ago
  16. jump Group Title
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    Thanksfor the extra effort to show me ur work , i'mgoing to have to study that before my next test!

    • one year ago
  17. PhoenixFire Group Title
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    No problem, when I learnt it I had lots of trouble getting it right. Just keep practicing!

    • one year ago
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