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Inverting a 4x4 matrix

  1. Aug 16, 2003 #1
    here's what i've done
    -got transpose of A

    |-3 0 0 1 |
    | 1 -2 1 0 |
    | 2 -3 2 1 |
    | 1 2 -1 2 |


    the deteminant = 8 therefore (its's invertible)

    **the problem***


    |-3 0 0 1 | 1 0 0 0 |
    | 1 -2 1 0 | 0 1 0 0 |
    | 2 -3 2 1 | 0 0 1 0 |
    | 1 2 -1 2 | 0 0 0 1 |

    i've tried reducing this (even tried using adjoint method)
    but i keep getting a different answer

    could someone please post for me the row reductions to
    get the inverse to this thing i.e
    row2 -row3 on row3 etc etc etc .
     
  2. jcsd
  3. Aug 16, 2003 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You talked about "reducing this" but didn't say exactly what it is you want to arrive at. If I remember correctly you wanted to find the inverse of a matrix. Was it A or the transpose of A? If you want to find the inverse of the matrix A, I don't see any reason to work with the transpose.

    |-3 0 0 1 | 1 0 0 0 |
    | 1 -2 1 0 | 0 1 0 0 |
    | 2 -3 2 1 | 0 0 1 0 |
    | 1 2 -1 2 | 0 0 0 1 |

    If I were going to do this, the first thing I would do is swap the first two rows so I will have a 1 in the first row of the first column. Then (1) add 3 times that (new) first row to the (new) second row, (2) subtract 2 times that (new) first row from the third row, and (3) Subtract that (new) first row from the fourth row to get:
    [1 -2 1 0| 0 1 0 0 ]
    [0 -6 3 1| 1 3 0 0 ]
    [0 1-2 1| 0-2 1 0 ]
    [0 4 2 2| 0-1 0 1 ]

    Now swap the second and third rows to get a 1 in the "pivot position" (second row, second column). Add twice that new second row to the first row, add 6 time the new second row to the third row, and subtract 4 times that new second row to the fourth row to get:

    OOPs, gotta run! (Going to see "The Barber of Seville"!) You'll have to work it out yourself.
     
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