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bob bundy
2012-10-18 08:50:27

Bob

Deon588
2012-10-18 07:29:03

Yes that was also the next question in the exam paper I did, worked out perfectly at the end

bob bundy
2012-10-17 23:39:52

hi Deon588

I know the rule for multiplying two matrices.  I dareay you could devise a rule for three but it would be hard to remember.  So two at a time is simplest.

You shouldn't assume that matrices can be swopped around like numbers but associativity { (AB)C = A(BC) } does work.

Beware: commutativity doesn't.  { AB ≠ BA }

Did you try my practice suggestion?

and you should find that inverse A = E3.E2.E1

This would be a good check that you have inv A right and that your multiplying is OK.

Bob

bobbym
2012-10-17 22:10:27

Hi Deon588;

Always break a big problem down into smaller pieces that you know how to do. This top down design is very common in programming. You knew how to do 2 matrices.

Deon588
2012-10-17 22:04:47

Hi Bob thanks a lot for all the effort.  I had no idea how to multiply more than 2 matrices so I tried doing all 3 at once...

bob bundy
2012-10-17 04:12:02

and you should find that inverse A = E3.E2.E1

This would be a good check that you have inv A right and that your multiplying is OK.

Bob

bob bundy
2012-10-17 04:04:49

hi Deon588

Your inverses are correct, so then you need to do the multiplying.

(AB)C = A(BC) for matrices so you have a choice of which pair you multiply first.  But you must preserve order.

eg if AB = D you then do DC not CD.

I'm going to do these both ways in case I choose the way you didn't (if you see what mean)

or

Bob

zetafunc.
2012-10-17 03:30:11

Never mind, bobbym was faster.

zetafunc.
2012-10-17 03:29:47

You multiplied the matrices incorrectly

bobbym
2012-10-17 03:28:19

Hi;

Your mistake is in line 3. Check how you multiplied those matrices.

Deon588
2012-10-17 03:17:36

Hi Bobbym any idea where I made a mistake?
Thanks a lot

bobbym
2012-10-17 00:32:17

Hi;

I am getting:

which can be inverted.

Deon588
2012-10-17 00:17:54

Hi all.  I think i'm making a mistake somewhere as the matrix I get after multiplying doesn't seem to be invertible and the next question is to find A^-1.  Is this the right way to do this?