# Elementary Row Operations

• Jun 10th 2008, 07:32 AM
taurus
Elementary 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 AM
Mathstud28
Quote:

Originally Posted by taurus
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

-4R2+R3?

Am I understanding you correctly?
• Jun 10th 2008, 07:35 AM
Moo
Hello,

Quote:

Originally Posted by taurus
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

Do R3=R3-4R2 :confused:
• Jun 10th 2008, 07:48 AM
angel.white
Quote:

Originally Posted by taurus
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

R2:=R2+R1
AND R3:=R3+3R1

(note that you misadded the last term in row2)

$\displaystyle \left[\begin{array}{cccc} -1& 3& 2& -8\\ 0& 3& 3& \color{red}{-10}\\ 0& 12& a+6& b-24\\ \end{array}\right]$

Quote:

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

What are you trying to accomplish with this?
• Jun 10th 2008, 09:26 AM
taurus
ok yea.
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 AM
Mathstud28
Quote:

Originally Posted by taurus
ok yea.
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?

-5R3+R2
• Jun 10th 2008, 09:41 AM
Moo
Quote:

Originally Posted by Mathstud28
-5R3+R2

Hep hep

You can't do anything you want. It depends on the real purpose of these operations...
Furthermore, you have to say where you do this operation, is it added to R3 or R2 or what ?

(Wink)
• Jun 10th 2008, 09:56 AM
taurus
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 AM
Mathstud28
Quote:

Originally Posted by taurus
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?

If you have a systems of equations with two equations and three variables, there are an infinite amount of solutions, but these solutions follow a pattern.Have you covered this in class yet?
• Jun 10th 2008, 10:08 AM
taurus
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 AM
angel.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 AM
TheEmptySet
Quote:

Originally Posted by taurus
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?

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 AM
Mathstud28
Quote:

Originally Posted by TheEmptySet
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 $\displaystyle Ax=0$

Emptyset you have an astounding way of laying out your LaTeX and regular text so it is extremely aesthetically pleasing (Nod)
• Jun 10th 2008, 10:34 AM
TheEmptySet
(Sun)Thank you very much (Sun)

I love this place!!!!

It makes happy(Clapping)

Thank you MathStud