Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Matrices Question

  • 15-01-2013 4:06pm
    #1
    Registered Users, Registered Users 2 Posts: 1,494 ✭✭✭


    Can anybody help me with this question?

    Question_zpsa95a7c58.jpg

    I know how to get the determinant but I'm not sure which of the 2 matrices I'm supposed to use for the question.

    thanks for any help


Comments

  • Moderators, Science, Health & Environment Moderators Posts: 1,852 Mod ✭✭✭✭Michael Collins


    The cofactors in the 2nd matrix are obtained by eliminating the row and column of the position in question from the first matrix, and obtaining the determinant of the reduced matrix left over (and then adjusting sign accordingly). For example, the first entry in the 2nd matrix (the matrix of cofactors) is -16. This is obtained by eliminating the first row and column from the first matrix, now you are left with the following

    [latex] a_{11} = +{\left|\matrix{-1 & 2 \cr 6 & 4 \cr} \right|} = -16. [/latex]

    Similarly for the second entry in the matrix of cofactors

    [latex] a_{12} = -{\left|\matrix{4 & 2 \cr -2 & 4 \cr} \right|} = -20. [/latex]

    Note the minus above. This alternates between plus and minus for each cofactor - if you like you can use the subscript to tell you if there needs to be an extra minus added, if the indices add to an odd number e.g. [latex] a_{12} \to [/latex] 1+2=3 above, then there needs to be an extra minus.

    This should allow you to figure out p, q and r. Come back with their values when you have them, the inverse part is very easy from that point on.


  • Registered Users, Registered Users 2 Posts: 1,494 ✭✭✭The_Gatsby


    Thanks very much, just worked it out and am delighted as this question will give me full marks in an exam tomorrow! Thanks a lot for the reply


Advertisement