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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw:1348860718372:dw|
i already did it ,but i want to know how to do it with a systematic way,thanks
ABAAB

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I am not sure but looks worth try.
\[ ABAAB = \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} \]
Is this matrices??-
yes
A.B.A.A.B=M for sure
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}}
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.
Indeed it workds ...i had been using * where i should have used .
woah
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 \[ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a +c & b +c\\ c & d \end{bmatrix} \]
woops!! that's b+d
\[ M = \begin{bmatrix} 3 & 4 \\ 3 + 2 & 4+3 \end{bmatrix} \] so you need to preceed the left to right by 1
Try to write the expressions for multiplying this matrices from another side.
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
add down {{1, 1}, {2, 3}} add up {{3, 4}, {2, 3}} add down {{3, 4}, {5, 7}}
i did it in a similar way but without using equations too
well ... i hope there is better method.
|dw:1348863383528:dw|
too mantain the first line and add to second i have to multlyby A
A(BA.AB)
multply by right is a linear combinations of rows
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.
that is what i did i find first row
first row is two times row1A+row2B
AB put in first 2row1+row 2
haha .... no probs man!!

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