Results 1 to 2 of 2

Math Help - Gaus Jordan ELimination compared to cofactor inverse matrix

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    15

    Exclamation Gaus Jordan ELimination compared to cofactor inverse matrix

    Hey guys i am stuck with the gaus jordan method of the following matrix.

    A = 1 2 3
    4 5 6
    3 1 -2

    For the co factor i worked out the C transpose as:

    1/3 (-16 7 3
    26 -8 6
    -11 5 -3)

    The gaus jordan is suppoesed to agree with my final cofactor matrix but i cant seem to understand how to get there?? i have donw the row eliminations to get the diagonal as 1 and the rest as 0.


    Add (-4 * row1) to row2

    Add (-3 * row1) to row3

    Divide row2 by -3

    Add (5 * row2) to row3

    Divide row3 by -1

    Add (-2 * row3) to row2

    Add (-3 * row3) to row1

    Add (-2 * row2) to row1

    But now i don't understand what i have to do now?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Latkan View Post
    Hey guys i am stuck with the gaus jordan method of the following matrix.

    A = 1 2 3
    4 5 6
    3 1 -2

    For the co factor i worked out the C transpose as:

    1/3 (-16 7 3
    26 -8 6
    -11 5 -3)

    The gaus jordan is suppoesed to agree with my final cofactor matrix but i cant seem to understand how to get there?? i have donw the row eliminations to get the diagonal as 1 and the rest as 0.


    Add (-4 * row1) to row2

    Add (-3 * row1) to row3

    Divide row2 by -3

    Add (5 * row2) to row3

    Divide row3 by -1

    Add (-2 * row3) to row2

    Add (-3 * row3) to row1

    Add (-2 * row2) to row1

    But now i don't understand what i have to do now?
    With the Gauss–Jordan method, for each row operation that you perform on A, you are supposed to carry out the same row operation starting with the identity matrix. This is best done by writing the identity matrix alongside A, and then performing the row operations on both. The process starts like this:

    \begin{aligned}\begin{bmatrix}1&2&3&\vdots&1&0&0\\ 4&5&6&\vdots&0&1&0\\ 3&1&-2&\vdots&0&0&1\end{bmatrix} &\longrightarrow\begin{bmatrix}1&2&3&\vdots&1&0&0\  \ 0&-3&-6&\vdots&-4&1&0\\ 0&-5&-11&\vdots&-3&0&1\end{bmatrix}\\ <br />
&\longrightarrow\begin{bmatrix}1&2&3&\vdots&1&0&0\  \ 0&1&2&\vdots&\tfrac43&-\tfrac13&0\\ 0&-5&-11&\vdots&-3&0&1\end{bmatrix}.\end{aligned}

    By the time that the left side of the double matrix has become the identity, the right side should give you the inverse matrix for A, like this:

    \begin{bmatrix}1&0&0&\vdots&-\tfrac{16}3&\tfrac73&-1\\ 0&1&0&\vdots&\tfrac{26}3&-\tfrac{11}3&2\\ 0&0&1&\vdots&-\tfrac{11}3&\tfrac53&-1\end{bmatrix}.

    Finally, you should always multiply the inverse matrix by the original matrix to check that the product is the identity matrix. If it isn't, that tells you that you have made a mistake somewhere.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding an inverse matrix using Gauss-Jordan
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: September 25th 2011, 12:28 PM
  2. Replies: 1
    Last Post: September 12th 2011, 09:03 PM
  3. Replies: 1
    Last Post: February 16th 2011, 02:06 PM
  4. Help! Matrix, Gauss-Jordan elimination!
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 22nd 2008, 05:05 AM
  5. Gaus Elimination Help
    Posted in the Business Math Forum
    Replies: 1
    Last Post: September 22nd 2008, 08:27 PM

Search Tags


/mathhelpforum @mathhelpforum