Linear Algebra: Row Echelon Form

[wtfbbq]

Having some issues completely understanding what it is, and how to change a matrix into REF.

example from my HW (which I have answers to)

reduce to REF (not sure how to properly represent matrixes in text)
1 2 4
3 8 11

answer is

1 8/3 11/3
0 1 -1/2

I can see that they got the answer for the top row by switching R1 and R2, then dividing R1 by three. I then subtract the new bottom row and I get 2/3 -1/3 . I also tried doing it another way, and i got the proper bottom row, but my top row was wrong. can anyone help?

edit: i also realize this isn't quite "highly technical" but i'm going out on a limb to say that posting this in ATOT wouldn't help me that much. and i know this is something very simple for people who know linear algebra, but my teacher ran out of time during his lecture that started linear algebra, so he didn't cover any of this.

 

[Born2bwire]

Originally posted by: wtfbbq
Having some issues completely understanding what it is, and how to change a matrix into REF.

example from my HW (which I have answers to)

reduce to REF (not sure how to properly represent matrixes in text)
1 2 4
3 8 11

answer is

1 8/3 11/3
0 1 -1/2

I can see that they got the answer for the top row by switching R1 and R2, then dividing R1 by three. I then subtract the new bottom row and I get 2/3 -1/3 . I also tried doing it another way, and i got the proper bottom row, but my top row was wrong. can anyone help?

edit: i also realize this isn't quite "highly technical" but i'm going out on a limb to say that posting this in ATOT wouldn't help me that much. and i know this is something very simple for people who know linear algebra, but my teacher ran out of time during his lecture that started linear algebra, so he didn't cover any of this.

0 2/3 -1/3 --> 0 1 -1/2

 

[wtfbbq]

DOH, DUH! thanks.

 

[TheoPetro]

Its pretty easy. What you are trying to get is a matrix like,
1XXX
01XX
001X
0001

You get this through altering each row but without changing the equality.

To help with the homework problem you can divide R2 (Row 2) by 3. This would be (3/3) (8/3) (11/3). Giving a new R2 of 1 (8/3) (11/3). because you have a leading 1, R2 - R1 -> R2 (read as "row 1 minus row 2 yields a new row 2"). This puts a zero unter R1's 1 and keeps the REF. Now just alter R2 again to get a 1 after your zero and you have it. This will work with any sized matricies.

I dont know if you covered this but REF is useful when solving a system of equations for unknowns. If you have something like 2x+3y-4z=8, you bsaically treat each column of the matrix as a variable. So all the x terms would go in one column and so on. The matrix would be a 1x4 matrix looking like [2 3 4 8]. With a system of equations you would just put each successive equation on the next row. So if the previous equation and 3x+2y+z=9, 9x-8y+7z=1, were a system of equations your matrix would be,

x y z
|2 3 4 8|
|3 2 1 9|
|9 8 7 1|

To solve for your variables x, y, and z, you would put this matrix into REF. This would give you what z is equal to. You can then plug it into the row above it to get y and then plug y and z into R1 to solve for x.