1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear Algebra - Inverse of Matrices

  1. Oct 21, 2007 #1
    I have a problem in which I am given a 2-by-2 matrix A (0,8 / 5,7) which I am asked to write as a product of elementary matrices. Then I am asked write the inverse of A as a product of elementary matrices. So I turned A into reduced row echlon form using the following steps:

    -Switch R1 and R2
    -1/5 * R1
    -1/8 * R2
    - (-7/5) * R2 + R1 into R1

    So the elementary matrices are:
    (0,1 / 1,0) * (1/5,0 / 0,1) * (1,0 / 1/8,0) and (1,-7/5 / 0,-7/5)

    Right?

    Then the inverse of A as a product of elementary matrices is just the inverse of each of (1,-7/5 / 0,-7/5) * (1,0 / 1/8,0) * (1/5,0 / 0,1) and (0,1 / 1,0).

    Does this make sense? Because I keep getting the wrong answer and I don't know where I've gone wrong....
     
  2. jcsd
  3. Oct 21, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No. Since you did not change the second row in the last step, the bottom row of the last elementary matrix is "0 1" not"0 -7/5". The elementary matrix corresponding to a row-operation is the matrix derived by applying that row operation to the identity matrix. Sutracting 7/5 of the second row from the first row of the identity matrix gives (1, -7/5, 0, 1).

    The last elementary matrix is (1 , -7/5, 0, 1) and its inverse is (1, +7/5, 0, 1).
     
  4. Oct 21, 2007 #3
    Thanks! I actually just figured it out by myself all it took was a couple hours staring at it blankly before something clicked in my brain! :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear Algebra - Inverse of Matrices
Loading...