Results 1 to 4 of 4

Math Help - Augmented Matrix Problem

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    2

    Augmented Matrix Problem

    Hello, I have tried this problem out but get stuck on one part.

    Find an equation involving g, h and k that makes this augmented matrix consistent.

    1 -4 7 g
    0 3 -5 h
    -2 5 -9 k

    so i try and put this in trangular form by multiplying row_1 by 2 and adding it to row_3 and end up with
    1 -4 7 g
    0 3 -5 h
    0 -3 5 k
    can i assume since row 2 and row 3 are almost the same, just scaled by -1 that i can take out one of the equations since they are the same? would column 3 be a free variable?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Your conclusions are good, but your addition needs some work.

    The 'k' in the second tableau should be "k + 2g"
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2007
    Posts
    2
    ahhh yes, thanks, i forgot to look at the right side of the matrix.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, carlo_b!

    Find an equation involving g, h and k that makes this augmented matrix consistent.

    . . \begin{pmatrix}1 & \text{-}4 & 7 &|&  g \\<br />
0 &  3 & \text{-}5 &|&  h \\<br />
\text{-}2 &  5 & \text{-}9 &|&  k \end{pmatrix}

    Your next step is a bit off . . .

    \begin{array}{c} \\ \\ R_3 + 2\!\cdot\!R_1\end{array}\;\begin{pmatrix}1 & \text{-}4 & 7 &|&  g \\<br />
0 & 3 & \text{-}5 &|& h \\<br />
0 & \text{-}3 & 5 &|& {\color{blue}k+2g} \end{pmatrix}


    Can i assume since row 2 and row 3 are almost the same, just scaled by -1,
    that i can take out one of the equations since they are the same? . . Yes
    Would column 3 be a free variable? . Yes
    One more step . . .

    \begin{array}{c}\\ \\ \text{-}1\!\cdot\!R_3\end{array}\;\begin{pmatrix}1 & \text{-}4 & 7 &|& g \\ 0 & 3 & \text{-}5 &|& h \\ 0 & 3 & \text{-}5 &|& \text{-}2g-k\end{pmatrix}


    If h is not equal to \text{-}2g-k, the system is inconsistent.


    To be consistent: . h \;=\;\text{-}2g-k



    Ahh, too slow again . . .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: September 1st 2011, 04:35 AM
  2. Augmented matrix
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 3rd 2009, 10:01 AM
  3. Augmented matrix
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 9th 2008, 08:17 PM
  4. Augmented matrix
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 31st 2007, 07:39 AM
  5. Augmented matrix
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 19th 2006, 11:10 AM

Search Tags


/mathhelpforum @mathhelpforum