## 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

1. 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 ?

2. anonymous

Got it?

3. anonymous

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

4. anonymous

oops, I meant 1 0 0 -11 -25 0 1 0 3 6 0 0 1 0 2 0 0 0 1 -2

5. anonymous

6. anonymous

ok, I redo it. let clear them all.

7. anonymous

I cleared mine!

8. anonymous

$$\left[\begin{matrix}1&3&0&-2&7\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]$$

9. 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?

10. anonymous

ok?

11. anonymous

No... Oh my... how did you end up with -11 as your row one column five answer? It used to be -25...

12. 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)

13. anonymous

ok??

14. 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.

15. anonymous

aaaaaaaaaaaaah, your last one is -7, not 7, ok hence it is -25. I am so so sorry

16. anonymous

$\left[\begin{matrix}1&0&0&-11&-25\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]$

17. anonymous

Step2: 11R4 + R1 to get new row 1

18. anonymous

So, multiply row 4 by 11 and add to row one to get the new row one? Why row 4?

19. anonymous

$\left[\begin{matrix}1&0&0&0&-45\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]$

20. 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.

21. 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?

22. anonymous

So, what is the next step?

23. 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.

24. anonymous

Ah, I see.

25. 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

26. anonymous

So the new row2 would be [0 1 0 0 0]?

27. anonymous

yes, and you are done.

28. anonymous

$\left[\begin{matrix}1&0&0&0&-45\\0&1&0&0&0\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]$

29. 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.

30. anonymous

yes, it is.

31. anonymous

unless the last column, all of the entries are either 1 or 0.

32. anonymous

with 1's are in diagonal line.

33. anonymous

Thank you so much! I really appreciate your help! I "fanned" you or whatever the follow thing on here is.

34. anonymous

ok, don't forget: I am crazy one.

35. anonymous

Crazy One?