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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348860718372:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i already did it ,but i want to know how to do it with a systematic way,thanks

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1I am not sure but looks worth try.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1\[ ABAAB = \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} \]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1hold 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}}

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1http://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.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1Indeed it workds ...i had been using * where i should have used .

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1let'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 \[ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a +c & b +c\\ c & d \end{bmatrix} \]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1\[ M = \begin{bmatrix} 3 & 4 \\ 3 + 2 & 4+3 \end{bmatrix} \] so you need to preceed the left to right by 1

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0Try to write the expressions for multiplying this matrices from another side.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1A 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

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1add down {{1, 1}, {2, 3}} add up {{3, 4}, {2, 3}} add down {{3, 4}, {5, 7}}

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i did it in a similar way but without using equations too

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1well ... i hope there is better method.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348863383528:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0too mantain the first line and add to second i have to multlyby A

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0multply by right is a linear combinations of rows

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1well ... 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is what i did i find first row

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first row is two times row1A+row2B

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0AB put in first 2row1+row 2

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1haha .... no probs man!!
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