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Matrix row reduction

  1. May 13, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the inverse of


    by reducing the augmented matrix [A|I] using row operations.

    3. The attempt at a solution

    I just can't seem to get this one out, it's the last in a series of 9 similar questions and the first I am to have a hard time with.
    First few attempts involved operations like:
    R1 <- R1 x 1/4
    R3 <- R3 + R2
    R4 <- R4 + R1
    R1 <- R1 - 2R4

    But I don't seem to be getting any closer to a solution with the operations that i am trying.

    Any help, even just a pointer as to what row operations to start off with, would be gratefully appreciated.
  2. jcsd
  3. May 13, 2007 #2

    D H

    Staff: Mentor

    You want to use row operations to transform the left half of the augmented matrix into the identity matrix. Do this in two stages: (1) transform the left half of the augmented matrix into upper-triangular form, and (2) transform this into the identity matrix. You can do the scaling so that the diagonals are all ones as part of the first stage or at the very end.

    Now split the first stage into substages. Use row operations to make the first column of rows 2-4 zero. Then make the second column of rows 3 and 4 zero. Finally, make the third column of row 4 zero.

    You started fine. Your second step, R3 <- R3 + R2, is where you started to go awry.
  4. May 13, 2007 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    I usually start these by switching row 3 with row 1, so I don't need to deal with fractions from step one (because you already have a 1 in slot 1,1). Note you can leave row 2 where it is and you already have a 1, so if you just kill the first two columns before going on to deal with columns three and four, you've already reduced the number of fractions you're dealing with by half
  5. May 13, 2007 #4
    thanks a lot guys.
    just got it out then, unsure of correctness, but i've gotten a result and that's at least worth part marks :p
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