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, 07:32 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, 07:33 AMMathstud28
- Jun 10th 2008, 07:35 AMMoo
- Jun 10th 2008, 07:48 AMangel.white
R2:=R2+R1

AND R3:=R3+3R1

(note that you misadded the last term in row2)

$\displaystyle \left[\begin{array}{cccc}

-1& 3& 2& -8\\

0& 3& 3& \color{red}{-10}\\

0& 12& a+6& b-24\\

\end{array}\right]$

What are you trying to accomplish with this? - Jun 10th 2008, 09:26 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, 09:37 AMMathstud28
- Jun 10th 2008, 09:41 AMMoo
- Jun 10th 2008, 09:56 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, 10:02 AMMathstud28
- Jun 10th 2008, 10:08 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, 10:08 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, 10:10 AMTheEmptySet
You can paramterize the soutuions as follows

let $\displaystyle z=t$

Then the equation above the row of zero's is

$\displaystyle -5y+10z=-15 \iff -5y+10(t)=-15 \iff y=3+2t$ and using these we can solve for x

$\displaystyle x+2(3+2t)-4(t)=10 \iff x+6+4t-4t=10 \iff x=4$

So finally we get the solution

$\displaystyle (4,3+2t,t)$

This can be broken up into the particular and complimentry solution

$\displaystyle (4,3+2t,t)=(4,3,0)+t(0,2,1)$

where $\displaystyle y_p=(4,3,0) \mbox{ and } y_c=t(0,2,1)$

This is true for any $\displaystyle t \in \mathbb{R}$

Note that $\displaystyle y_c$ spans the null space of the homogenious system $\displaystyle Ax=0$ - Jun 10th 2008, 10:24 AMMathstud28
- Jun 10th 2008, 10:34 AMTheEmptySet
(Sun)Thank you very much (Sun)

I love this place!!!!

It makes happy(Clapping)

Thank you MathStud