## anonymous 3 years ago write M as a product of many factors A and B.

1. anonymous

|dw:1348860718372:dw|

2. anonymous

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

3. anonymous

ABAAB

4. experimentX

I am not sure but looks worth try.

5. experimentX

$ABAAB = \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$

6. anonymous

Is this matrices??-

7. anonymous

yes

8. anonymous

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

9. experimentX

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

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

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

12. anonymous

woah

13. experimentX

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

woops!! that's b+d

15. experimentX

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

16. klimenkov

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

17. experimentX

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

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

19. anonymous

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

20. experimentX

well ... i hope there is better method.

21. anonymous

|dw:1348863383528:dw|

22. anonymous

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

23. anonymous

A(BA.AB)

24. anonymous

multply by right is a linear combinations of rows

25. experimentX

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. anonymous

that is what i did i find first row

27. anonymous

first row is two times row1A+row2B

28. anonymous

AB put in first 2row1+row 2

29. anonymous

thanks @experimentX

30. experimentX

haha .... no probs man!!