Re: inverse question...


[ Follow Ups ] [ Post Followup ] [ Calculus Message Board ] [ FAQ ]

Posted by Soroban on September 29, 2002 at 13:05:16:

In Reply to: inverse question... posted by jessica on September 28, 2002 at 23:38:21:

: find A^-1 or A inverse:

: 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

Hello, Jessica!

Yes, your procedure is absolutely correct.

To eliminate the -1 at (3,1), I would multiply the row-1 by i, and add it to row-3.
(I don't change row-1; I do the "times i, add to row-3" on the side.)
I also note it as "III + iI" next to row-3. (It works for me!)
Then the new row-3 is: | 0 -i 2 | 0 0 1 |.

We move to the second column (reading down): 1+i, 1, -i
Then, using my notation: I - (1+i)II, and III + iII

Then clear out the third column in the same manner.

I hope you can follow this "code" I use.


Follow Ups:



Post a Followup

Name:
E-Mail:

Subject:

Comments:

Optional Link URL:
Link Title:
Optional Image URL:


[ Follow Ups ] [ Post Followup ] [ Calculus Message Board ] [ FAQ ]