Re: Fact about 3x3 Matrices (Nicole)

[James Crook]

Thanks to Nicole's post we know how to find the inverse of:

5 0 0
0 3 0
0 0 8

We'll soon be able to find the inverse of:

cos(t)  0   sin(-t)
0        2          0
sin(t)   0   cos(t)

No one has explained how to do this one yet, so I think it will help to do some simpler matrix examples first.

So here are two problems about 2x2 matrices.

Anyone can step in and explain and offer an answer to one or the other or both.

1) I'd like to know what the determinant of this matrix is:

6  7

4  5

2) I'd like an example of two 2x2 matrices A and B where A x B is not the same as B x A.  (where 'x' is matrix multiplication).

This is not like ordinary multiplication where 3 x 4 is the same thing as 4 x 3.

--James.

[Jonathan Joo]

The determinant of the 2x2 matrix is 2, I believe. (30-28)

 

As for an example of a matrix,

 

2 3    and    1 2

4 5              2 3

 

work, but obviously there are many more matricies that satisfy that condition.

 

Sorry I haven't thought of a fact yet, I'll get back to you soon on that!

 

-Jonathan

[James Crook]

Yes, those are correct.

The determinant is 6x5 - 4x7 = 2.

And those two matrices do not commute - and as you say there are loads of examples that don't.

Looking forward to the fact.

---

Would someone else like to give an example of two different 2x2 matrices which do commute, ie. AxB = BxA with no zeroes in the matrices, and explain for the group how they chose them?  [Yes, I'm not letting you use the identity matrix for one of them, that would be just too easy].

--James.

[Nicole Shih]

If the matrices all contain the same numbers like

[2  2] [3  3] then the resulting matrices will always be [12  12]

[2  2] [3  3]                                                                   [12  12]

[James Crook]

Great Nicole.

In this example one matrix is a multiple of another.  That generalizes.

If we are given any matrix A, perhaps 

6 7

4 5 

then we know kA will commute with A.    

so 2A, which is:

12 14

 8 10

will commute with A.  I make the product to be:

128 154

 88  106

both for AB and for BA.  Would someone check my arithmetic?

Can we find another matrix B that commutes with A, so AB=BA, where B isn't a constant multiple of A?  Once again we know the Identity matrix I and multiples of it will commute with A, AI=A=IA, so I don't want I or multiples of it.

This time I'd like an answer that is not from Jonathan and not from Nicole

[Kelsey]

If we have a matrix A such as

1 3
2 4

then its inverse, B, is

-2 3/2
1 -1/2

and AB = BA = I.

[James Crook]

Yes.

A^-1 commutes with A.  And because of associativity we can show A^2 commutes with A and A^3, and A^-2 and so on.

Can we use distributivity to get any other possibilites?  And if we can, have we got them all??