Question

Solve the following system of linear equations using elementary row operations:

2x + 7y +1z = -2
x + 4y + 3z = -2
y + 5z = -4

Answer 1

Do you know how to put that in matrix form first? That'll be your first step.

Answer 2

Sure:

[ 0 1 5 -4
1 4 3 -2
2 7 1 -2 ]

Answer 3

I'm just blanking on the next step.

Answer 4

Do you want to go to row-echelon form or reduced-row-echelon form?

Answer 5

Haven't gotten that far yet, I'm just doing row replacement and scaling.

Answer 6

Either way, a good first step is to swap the [1 4 3 -2] row to the top so you'll already have a leading 1 in the upper left corner.

Answer 7

I tried replacing [2 7 1 -2] with 2* [1 4 3 -2} and adding it to the [2 7 1 -2] row, giving me [ 0 -1 -5 2], but I don't think that's right.

Answer 8

That will also conveniently put the [0 1 5 -4] row under it, so you'll have a 0 under the first leading 1 and the next leading 1 in the right place.

Answer 9

...which will also put 2 7 1 -2 in the 3rd row, putting the 1 in the correct place there.

Answer 10

I'm starting with this matrix:
[1 4 3 -2]
[0 1 5 -4]
[2 7 1 -2]

If you want to go all the way to reduced-row-echelon form (which gives all the solutions with no further work), then your goal is to get the matrix to look like this:
[1 0 0 x]
[0 1 0 y]
[0 0 1 z]

Answer 11

ah, that's what i want to do, we haven't gone over the terminology yet.

Answer 12

So the best way to go is to go column by column getting zeroes above and below the leading 1's.

Answer 13

Oh!

Answer 14

So if I add the sum of -2*row 1 and row 3 to replace row 3, i'll have all zeroes in the first column?

Answer 15

The best way to do that is to use multiples of the leading 1 and subtract that from what you're trying to get rid of.
e.g. subtract twice row 1 from row three...

Answer 16

Read my mind! Exactly.

Answer 17

Ok, I think I've got it from here, thanks for the help!

Answer 18

y.w. Enjoy!