Results 1 to 14 of 14

Math Help - Elementary Row Operations

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    222

    Question 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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by taurus View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Quote Originally Posted by taurus View Post
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    Quote Originally Posted by taurus View Post
    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)

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

    Quote Originally Posted by taurus View Post
    But now am stuck, how do i get a 0 in the position of 12?
    What are you trying to accomplish with this?
    Last edited by angel.white; June 10th 2008 at 08:52 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2007
    Posts
    222

    Question

    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by taurus View Post
    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?
    -5R3+R2
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Mathstud28 View Post
    -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 ?

    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Sep 2007
    Posts
    222

    Question

    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?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by taurus View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Sep 2007
    Posts
    222

    Question

    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by taurus View Post
    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 z=t

    Then the equation above the row of zero's is

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

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

    So finally we get the solution

    (4,3+2t,t)

    This can be broken up into the particular and complimentry solution


    (4,3+2t,t)=(4,3,0)+t(0,2,1)

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

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

    Note that y_c spans the null space of the homogenious system Ax=0
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by TheEmptySet View Post
    You can paramterize the soutuions as follows

    let z=t

    Then the equation above the row of zero's is

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

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

    So finally we get the solution

    (4,3+2t,t)

    This can be broken up into the particular and complimentry solution


    (4,3+2t,t)=(4,3,0)+t(0,2,1)

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

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

    Note that y_c spans the null space of the homogenious system Ax=0
    Emptyset you have an astounding way of laying out your LaTeX and regular text so it is extremely aesthetically pleasing
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Thank you very much

    I love this place!!!!

    It makes happy

    Thank you MathStud
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Algorithm and elementary operations
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 30th 2011, 08:25 AM
  2. Elementary Row Operations
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: October 9th 2010, 03:44 AM
  3. Elementary row operations
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 1st 2009, 09:23 PM
  4. Elementary row operations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 17th 2007, 08:01 AM
  5. Elementary Row Operations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 17th 2007, 07:59 AM

Search Tags


/mathhelpforum @mathhelpforum