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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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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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 ?
Got it?
next is -3 R3 + R2 to get a new row 2

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

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