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RaphaelFilgueiras

write M as a product of many factors A and B.

  • one year ago
  • one year ago

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  1. RaphaelFilgueiras
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    |dw:1348860718372:dw|

    • one year ago
  2. RaphaelFilgueiras
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    i already did it ,but i want to know how to do it with a systematic way,thanks

    • one year ago
  3. RaphaelFilgueiras
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    ABAAB

    • one year ago
  4. experimentX
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    I am not sure but looks worth try.

    • one year ago
  5. experimentX
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    \[ ABAAB = \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} \]

    • one year ago
  6. bthemesandtricks
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    Is this matrices??-

    • one year ago
  7. RaphaelFilgueiras
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    yes

    • one year ago
  8. RaphaelFilgueiras
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    A.B.A.A.B=M for sure

    • one year ago
  9. experimentX
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    hold on ... looks I'm doing wrong calculation with MMA http://www.wolframalpha.com/input/?i={{1%2C+0}%2C+{1%2C+1}}*{{a%2C+b}%2C+{c%2C+d}}

    • one year ago
  10. experimentX
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    http://www.wolframalpha.com/input/?i={{1%2C+1}%2C+{0%2C+1}}*{{a%2C+b}%2C+{c%2C+d}} let's begin with the nature of those transformations.

    • one year ago
  11. experimentX
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    Indeed it workds ...i had been using * where i should have used .

    • one year ago
  12. bthemesandtricks
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    woah

    • one year ago
  13. experimentX
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    let's begin with the nature of transformation \[ \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & b \\ a+ c & b+d \end{bmatrix} \] and \[ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a +c & b +c\\ c & d \end{bmatrix} \]

    • one year ago
  14. experimentX
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    woops!! that's b+d

    • one year ago
  15. experimentX
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    \[ M = \begin{bmatrix} 3 & 4 \\ 3 + 2 & 4+3 \end{bmatrix} \] so you need to preceed the left to right by 1

    • one year ago
  16. klimenkov
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    Try to write the expressions for multiplying this matrices from another side.

    • one year ago
  17. experimentX
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    A adds to the bottom while B adds to top. since 4>3 let's looks for B.A = {{2, 1}, {1, 1}} A.B = {{1, 1}, {1, 2}} so A.B is our choice

    • one year ago
  18. experimentX
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    add down {{1, 1}, {2, 3}} add up {{3, 4}, {2, 3}} add down {{3, 4}, {5, 7}}

    • one year ago
  19. RaphaelFilgueiras
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    i did it in a similar way but without using equations too

    • one year ago
  20. experimentX
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    well ... i hope there is better method.

    • one year ago
  21. RaphaelFilgueiras
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    |dw:1348863383528:dw|

    • one year ago
  22. RaphaelFilgueiras
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    too mantain the first line and add to second i have to multlyby A

    • one year ago
  23. RaphaelFilgueiras
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    A(BA.AB)

    • one year ago
  24. RaphaelFilgueiras
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    multply by right is a linear combinations of rows

    • one year ago
  25. experimentX
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    well ... i kept fitting up. if you intoduce difference and use that difference to introduce another difference .. then it would be (2d) double step. so keep adding from 1, 1 (no difference) to 1 2 .... until the difference is able to give exact value at top. and use that difference to create double difference at bottom.

    • one year ago
  26. RaphaelFilgueiras
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    that is what i did i find first row

    • one year ago
  27. RaphaelFilgueiras
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    first row is two times row1A+row2B

    • one year ago
  28. RaphaelFilgueiras
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    AB put in first 2row1+row 2

    • one year ago
  29. RaphaelFilgueiras
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    thanks @experimentX

    • one year ago
  30. experimentX
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    haha .... no probs man!!

    • one year ago
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