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anonymous
 one year ago
Thee augmented matrix of a linear system has been reduced by row operations to the form shown. in each case, continue the appropriate row operations and describe the solution set of the original system
The matrix is
1 3 0 2 7
0 1 0 3 6
0 0 1 0 2
0 0 0 1 2
anonymous
 one year ago
Thee augmented matrix of a linear system has been reduced by row operations to the form shown. in each case, continue the appropriate row operations and describe the solution set of the original system The matrix is 1 3 0 2 7 0 1 0 3 6 0 0 1 0 2 0 0 0 1 2

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is reduce echelon form already. To get the row reduce echelon form from here, you need: 3R2 + R1 to get rid off 3 on entry \(A_{1,2}\) . Show me what you get ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0next is 3 R3 + R2 to get a new row 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, I meant 1 0 0 11 25 0 1 0 3 6 0 0 1 0 2 0 0 0 1 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, I redo it. let clear them all.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\left[\begin{matrix}1&3&0&2&7\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&2\end{matrix}\right]\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01 0 0 11 25 (after 3R2 + R1) 0 1 3 3 0 (after 3R3 + R2) 0 0 1 0 2 0 0 0 1 2 Is that correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No... Oh my... how did you end up with 11 as your row one column five answer? It used to be 25...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0step1 it is 3R2 + R1 right? 3* (0 1 0 3 6) = (0 3 0 9 18) + Row1 (1 3 0 2 7) that is  new row 1: (1 0 0 11 11)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh my ^_^; I see what the problem is. The seven at the end of row one is a 7 in the problem (a simple typo. 7 +  18 = 25.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0aaaaaaaaaaaaah, your last one is 7, not 7, ok hence it is 25. I am so so sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}1&0&0&11&25\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&2\end{matrix}\right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Step2: 11R4 + R1 to get new row 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, multiply row 4 by 11 and add to row one to get the new row one? Why row 4?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}1&0&0&0&45\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&2\end{matrix}\right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good question: Why R4, because R4 has all entry before 1 is 0's, hence when you multiple to any number, it doesn't change other entry.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah! I see. Good point. So when I go through this process I should choose the row that is the closest and/or most convenient for the row I am trying to change?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, what is the next step?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}1&0&0&\color{red}{11}&25\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&2\end{matrix}\right]\] I choose the row 4, because I want to get rid of the number right in above its 1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To the very brand new matrix, \[\left[\begin{matrix}1&0&0&0&45\\0&1&0&\color{red}{3}&6\\0&0&1&0&2\\0&0&0&1&2\end{matrix}\right]\] This guy is the number I want to get rid off, and it stay right above 1 of R4, right? Hence again, 3 R4+R2 to get new R2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the new row2 would be [0 1 0 0 0]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, and you are done.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}1&0&0&0&45\\0&1&0&0&0\\0&0&1&0&2\\0&0&0&1&2\end{matrix}\right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So wait... The entire goal of this problem is basically to get all of the columns before the last one to be filled with ones and zeros.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0unless the last column, all of the entries are either 1 or 0.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0with 1's are in diagonal line.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much! I really appreciate your help! I "fanned" you or whatever the follow thing on here is.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, don't forget: I am crazy one.
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