anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
That 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
  • anonymous
Got it?
anonymous
  • anonymous
next is -3 R3 + R2 to get a new row 2

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anonymous
  • anonymous
oops, I meant 1 0 0 -11 -25 0 1 0 3 6 0 0 1 0 2 0 0 0 1 -2
anonymous
  • anonymous
oh, gosh, my bad,
anonymous
  • anonymous
ok, I redo it. let clear them all.
anonymous
  • anonymous
I cleared mine!
anonymous
  • anonymous
\(\left[\begin{matrix}1&3&0&-2&7\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\)
anonymous
  • anonymous
1 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
  • anonymous
ok?
anonymous
  • anonymous
No... Oh my... how did you end up with -11 as your row one column five answer? It used to be -25...
anonymous
  • anonymous
step1 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
  • anonymous
ok??
anonymous
  • anonymous
Oh 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
  • anonymous
aaaaaaaaaaaaah, your last one is -7, not 7, ok hence it is -25. I am so so sorry
anonymous
  • anonymous
\[\left[\begin{matrix}1&0&0&-11&-25\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\]
anonymous
  • anonymous
Step2: 11R4 + R1 to get new row 1
anonymous
  • anonymous
So, multiply row 4 by 11 and add to row one to get the new row one? Why row 4?
anonymous
  • anonymous
\[\left[\begin{matrix}1&0&0&0&-45\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\]
anonymous
  • anonymous
Good 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
  • anonymous
Ah! 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
  • anonymous
So, what is the next step?
anonymous
  • anonymous
\[\left[\begin{matrix}1&0&0&\color{red}{-11}&-25\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\] I choose the row 4, because I want to get rid of the number right in above its 1.
anonymous
  • anonymous
Ah, I see.
anonymous
  • anonymous
To 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
  • anonymous
So the new row2 would be [0 1 0 0 0]?
anonymous
  • anonymous
yes, and you are done.
anonymous
  • anonymous
\[\left[\begin{matrix}1&0&0&0&-45\\0&1&0&0&0\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\]
anonymous
  • anonymous
So 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
  • anonymous
yes, it is.
anonymous
  • anonymous
unless the last column, all of the entries are either 1 or 0.
anonymous
  • anonymous
with 1's are in diagonal line.
anonymous
  • anonymous
Thank you so much! I really appreciate your help! I "fanned" you or whatever the follow thing on here is.
anonymous
  • anonymous
ok, don't forget: I am crazy one.
anonymous
  • anonymous
Crazy One?

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