Question

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

Answer 1

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 ?

Answer 2

Got it?

Answer 3

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

Answer 4

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

Answer 5

oh, gosh, my bad,

Answer 6

ok, I redo it. let clear them all.

Answer 7

I cleared mine!

Answer 8

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

Answer 9

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?

Answer 10

ok?

Answer 11

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

Answer 12

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)

Answer 13

ok??

Answer 14

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.

Answer 15

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

Answer 16

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

Answer 17

Step2: 11R4 + R1 to get new row 1

Answer 18

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

Answer 19

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

Answer 20

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.

Answer 21

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?

Answer 22

So, what is the next step?

Answer 23

\[\left[\begin{matrix}1&0&0&\color{red}{-11}&-25\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\]
I choose the row 4, because I want to get rid of the number right in above its 1.

Answer 24

Ah, I see.

Answer 25

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

Answer 26

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

Answer 27

yes, and you are done.

Answer 28

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

Answer 29

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.

Answer 30

yes, it is.

Answer 31

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

Answer 32

with 1's are in diagonal line.

Answer 33

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

Answer 34

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

Answer 35

Crazy One?