# Re: matrix inverse...

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Posted by T.Gracken on September 29, 2002 at 06:51:12:

In Reply to: matrix inverse... posted by jessica on September 28, 2002 at 23:02:44:

: find A^-1 or A inverse:

: A= 1 1+i 0
: 0 1 i
: -i 1-2i 2

: ...i know the first step is to set it up like this:

: 1 1+i 0|1 0 0
: 0 1 i|0 1 0
: -i 1-2i 2|0 0 1

: and then try to make the first section be the identity matrix and the matrix next to it would be the inverse. However, i am not quite sure how to isolate the -i on the bottom first row as well as the other 'a+bi' terms...thanks for the help

don't let the "i" mess with you.

You already have a 1 in row-1, col-1. So next is zeros below it. row-2 already has 0, so for row three, use the row operation that allows multiplying a row by a number and adding it to another row. that is, multiply row 1 by i and add to row-3

you should now have
(look in the comments box below to see the matrix with columns lined up)

: 1 1+i 0 | 1 0 0
: 0 1 i | 0 1 0
: 0 2 2 | i 0 1

Now, you also have a 1 in row-2,col-2 so get zeros above and below... for row-1 multiply row-2 by (1+i) and subtract from row-1. for row-3, multiply row-2 by 2 and subtract from row-3,

once that is done,multiply row-3 by reciprocal of number in row-3,col-3. then get zeros again and you'll be done.

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