Advertisement
Help Keep Boards Alive. Support us by going ad free today. See here: https://subscriptions.boards.ie/.
If we do not hit our goal we will be forced to close the site.

Current status: https://keepboardsalive.com/

Annual subs are best for most impact. If you are still undecided on going Ad Free - you can also donate using the Paypal Donate option. All contribution helps. Thank you.
https://www.boards.ie/group/1878-subscribers-forum

Private Group for paid up members of Boards.ie. Join the club.

.: Matrices Help

  • 17-04-2008 12:04PM
    #1
    eoiner-galway
    Closed Accounts Posts: 73 ✭✭


    I was wondering if anyone can help me with this question
    1 0 2
    the matrix A = 0 1 -4
    2 1 3 cant do brackets, sry

    a) compute AtA, where At denotes the transpose of A

    b) Compute the determinant of A

    c) Find A(to the power)-1 and use it to solve the following system of equations
    x+2z= -1
    y-4z= -1
    2x+y+3z= 2
    thanks for the help


Comments

  • #2
    L'prof
    Registered Users, Registered Users 2 Posts: 12,053 ✭✭✭✭L'prof


    eoiner-galway wrote: »
    I was wondering if anyone can help me with this question
    1 0 2
    the matrix A = 0 1 -4
    2 1 3 cant do brackets, sry

    a) compute AtA, where At denotes the transpose of A

    b) Compute the determinant of A

    c) Find A(to the power)-1 and use it to solve the following system of equations
    x+2z= -1
    y-4z= -1
    2x+y+3z= 2
    thanks for the help

    So A =
    1 0 2
    0 1 -4
    2 1 3

    For the transpose of A, you simply swap the rows for the columns! So, first row of A becomes the first column of At and so on!

    AtA you multiply the two matrices!

    Determinant of A, couple of ways to do this! Here's one of them!

    I presume you mean A inverse? 1/(determinant) * transpose!

    This should be enough to solve the system of equations!


  • #3
    eoiner-galway
    Closed Accounts Posts: 73 ✭✭eoiner-galway


    yeh, sorry about the postitioning... it looked inline before i posted it


  • #4
    Tar.Aldarion
    Moderators, Science, Health & Environment Moderators, Social & Fun Moderators, Society & Culture Moderators Posts: 60,119 Mod ✭✭✭✭Tar.Aldarion


    There is another thread on matrices here at the moment that should help you. Also there are a lot of sites etc that will tell you what to do.


  • #5
    Dufresne
    Registered Users, Registered Users 2 Posts: 201 ✭✭Dufresne


    AtA is
    5 2 8
    2 2 -1
    8 -1 29

    Det A is 3

    Inverse of A is
    7/3 2/3 -2/3
    -8/3 -1/3 4/3
    -2/3 -1/3 1/3

    Good luck with the rest


  • #6
    eoiner-galway
    Closed Accounts Posts: 73 ✭✭eoiner-galway


    thanks, the rest is the part i need though...

    thanks anyway


  • Advertisement
  • #7
    Fremen
    Registered Users, Registered Users 2 Posts: 2,481 ✭✭✭Fremen


    Any textbook will tell you how to do this stuff, it's practically the first thing you learn in a linear algebra course. Go to the library and look up "solving systems of linear equations".


  • #8
    cunnins4
    Closed Accounts Posts: 882 ✭✭✭cunnins4


    Let A=

    1 0 2
    0 1 -4
    2 1 3

    Let B=

    x
    y
    z

    and C=

    1
    -1
    2

    inverse(A)*A = I, the identity matrix yeah? And I*B=B.

    If you have the inverse of A, you can use the inverse of A to find x y and z.

    A*B=C
    A*inverse(A)*B=inverse(A)*C

    therefore

    I*B=inverse(A)*C

    so

    B=inverse(A)*C

    I think that's correct anyways!:p


Advertisement