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

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

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Other answers:

I am not sure but looks worth try.
\[ ABAAB = \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} \]
Is this matrices??-
A.B.A.A.B=M for sure
hold on ... looks I'm doing wrong calculation with MMA{{1%2C+0}%2C+{1%2C+1}}*{{a%2C+b}%2C+{c%2C+d}}{{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 .
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.
too mantain the first line and add to second i have to multlyby A
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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