Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

RaphaelFilgueiras

  • 3 years ago

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

  • This Question is Closed
  1. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1348860718372:dw|

  2. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i already did it ,but i want to know how to do it with a systematic way,thanks

  3. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ABAAB

  4. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I am not sure but looks worth try.

  5. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[ ABAAB = \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} \]

  6. bthemesandtricks
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is this matrices??-

  7. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes

  8. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    A.B.A.A.B=M for sure

  9. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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}}

  10. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  11. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Indeed it workds ...i had been using * where i should have used .

  12. bthemesandtricks
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    woah

  13. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \[ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a +c & b +c\\ c & d \end{bmatrix} \]

  14. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    woops!! that's b+d

  15. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[ M = \begin{bmatrix} 3 & 4 \\ 3 + 2 & 4+3 \end{bmatrix} \] so you need to preceed the left to right by 1

  16. klimenkov
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Try to write the expressions for multiplying this matrices from another side.

  17. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  18. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    add down {{1, 1}, {2, 3}} add up {{3, 4}, {2, 3}} add down {{3, 4}, {5, 7}}

  19. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i did it in a similar way but without using equations too

  20. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    well ... i hope there is better method.

  21. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1348863383528:dw|

  22. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    too mantain the first line and add to second i have to multlyby A

  23. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    A(BA.AB)

  24. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    multply by right is a linear combinations of rows

  25. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  26. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that is what i did i find first row

  27. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    first row is two times row1A+row2B

  28. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    AB put in first 2row1+row 2

  29. RaphaelFilgueiras
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thanks @experimentX

  30. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    haha .... no probs man!!

  31. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy