I got the following matrix:

-1 3 2 -8

1 0 1 -2

3 3 a b

Then i applied R2=R2+R1 AND R3=R3+3R1 which gives

-1 3 2 -8

0 3 3 -6

0 12 a+6 b-24

But now am stuck, how do i get a 0 in the position of 12?

Thanks

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- Jun 10th 2008, 08:32 AMtaurusElementary Row Operations
I got the following matrix:

-1 3 2 -8

1 0 1 -2

3 3 a b

Then i applied R2=R2+R1 AND R3=R3+3R1 which gives

-1 3 2 -8

0 3 3 -6

0 12 a+6 b-24

But now am stuck, how do i get a 0 in the position of 12?

Thanks - Jun 10th 2008, 08:33 AMMathstud28
- Jun 10th 2008, 08:35 AMMoo
- Jun 10th 2008, 08:48 AMangel.white
- Jun 10th 2008, 10:26 AMtaurus
ok yea.

What about this one:

1 2 -4 10

2 -1 2 5

1 1 -2 7

I got it to this:

1 2 -4 10

0 -5 10 -15

0 -1 2 -3

How do i go on there? - Jun 10th 2008, 10:37 AMMathstud28
- Jun 10th 2008, 10:41 AMMoo
- Jun 10th 2008, 10:56 AMtaurus
I reduced it and got:

1 2 -4 10

0 -5 10 -15

0 0 0 0

Now how would i find all the solutions? - Jun 10th 2008, 11:02 AMMathstud28
- Jun 10th 2008, 11:08 AMtaurus
yes i should have as i am doing a past exam paper and the question is there. am just not quite sure how to do it.

- Jun 10th 2008, 11:08 AMangel.white
Please post the original question as it appears in your assignment, I think that you have missed something which is making it extremely difficult to help you.

- Jun 10th 2008, 11:10 AMTheEmptySet
You can paramterize the soutuions as follows

let

Then the equation above the row of zero's is

and using these we can solve for x

So finally we get the solution

This can be broken up into the particular and complimentry solution

where

This is true for any

Note that spans the null space of the homogenious system - Jun 10th 2008, 11:24 AMMathstud28
- Jun 10th 2008, 11:34 AMTheEmptySet
(Sun)Thank you very much (Sun)

I love this place!!!!

It makes happy(Clapping)

Thank you MathStud