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RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
dw:1348860718372:dw
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
i already did it ,but i want to know how to do it with a systematic way,thanks
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
I am not sure but looks worth try.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ ABAAB = \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} \]
 one year ago

bthemesandtricksBest ResponseYou've already chosen the best response.0
Is this matrices??
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
A.B.A.A.B=M for sure
 one year ago

experimentXBest ResponseYou've already chosen the best response.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}}
 one year ago

experimentXBest ResponseYou've already chosen the best response.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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
Indeed it workds ...i had been using * where i should have used .
 one year ago

experimentXBest ResponseYou've already chosen the best response.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 \[ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a +c & b +c\\ c & d \end{bmatrix} \]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
woops!! that's b+d
 one year ago

experimentXBest 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
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Try to write the expressions for multiplying this matrices from another side.
 one year ago

experimentXBest ResponseYou've already chosen the best response.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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
add down {{1, 1}, {2, 3}} add up {{3, 4}, {2, 3}} add down {{3, 4}, {5, 7}}
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
i did it in a similar way but without using equations too
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
well ... i hope there is better method.
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
dw:1348863383528:dw
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
too mantain the first line and add to second i have to multlyby A
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
multply by right is a linear combinations of rows
 one year ago

experimentXBest ResponseYou've already chosen the best response.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.
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
that is what i did i find first row
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
first row is two times row1A+row2B
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
AB put in first 2row1+row 2
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
thanks @experimentX
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
haha .... no probs man!!
 one year ago
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