Homework Help: Find B such that AB=BA (Linear Alg)

1. May 17, 2012

QuarkCharmer

1. The problem statement, all variables and given/known data
This is from Lay, 2.1 #11 (the second part). Not a homework problem, just for fun.

Let A be the 3x3 matrix, A = [1,1,1; 1,2,3; 1,4,5]. Find a matrix B such that:
$$AB = BA$$
where B is not the zero or identity matrix

2. Relevant equations

3. The attempt at a solution

Okay, so I know that typically, AB != BA, since matrix multiplication is non-commutative, but in some cases it can happen. What I did was make some 3x3 matrix B:

B = [a,b,c; d,e,f; g,h,i]

Then I wrote out AB = BA in matrix form, and solved both sides. Let's call AB matrix C, and BA matrix D. I set each entry in C to it's corresponding entry in D to form a system of 9 equations and 9 unknowns. I turned this into a 9x10 augmented matrix and attempted to rref it with my TI-89 but the resulting matrix is too big to display on the screen. I can scroll right and left, but I can't see anything below row 6.

Now, I am pretty confident that, if I did find a value for each entry in B (a,b,c,...,i) then it would form a matrix B such that AB = BA. But I don't think this is the right way to go about this problem at all. I can't believe they would expect me to solve 9 equations with 9 unknowns.

So my questions are:

1.) Would my approach have worked, say, if I computed it in mathematica.

and

2.) What is the proper approach to tackle this problem? I know there must be some way to go about this that is reasonable.

Thanks!
-QC

Last edited: May 17, 2012
2. May 17, 2012

Dustinsfl

Why not just compute it be hand? You don't need software for a 3x3.

3. May 17, 2012

QuarkCharmer

It's a 9x10 (augmented),
I assumed B was 3x3. A and B both have 9 entries, so that's 9 equations. Unless there is some other way to go about it that I am not seeing.

4. May 17, 2012

If this:

2. Matrix Algebra

Introductory Example: Computer Models in Aircraft Design

2.1 Matrix Operations

2.2 The Inverse of a Matrix

2.3 Characterizations of Invertible Matrices

2.4 Partitioned Matrices

2.5 Matrix Factorizations

...

is the second chapter of your book then read section 2.2 & come back to the question

5. May 17, 2012

QuarkCharmer

That's the book (well the chapters). I guess I'll make note of it and skip it. Seems like they would ask questions possible with only the knowledge provided in chapters 1-2.1.

Would my method have worked, if I cared to work out the 9x9 matrix?

6. May 17, 2012

Steely Dan

Note that if you were to write this matrix equation, it would be three equations, not nine. Therefore the system is under-determined, and there may be many possible solutions. I'm not sure how you got nine equations out of it; they can't be unique.

7. May 17, 2012

CornMuffin

your method would work, and it would find all possible matrices for B, though I certainly wouldn't want to go through all that work

You could just forget about the values of A, and think about what matrices would make AB=BA, you have the zero matrix and the identity, but what other simple matrices can you come up with?

You could also just start off with a 2x2 matrix and see if it gives you some insight.

8. May 17, 2012

QuarkCharmer

I can't think of any way to form this as a 3x3 matrix.

Here is how I get 9x9.

The equations on the right are the simplified equations from the left. Then the standard matrix formed by the 9 equations is at the bottom. In this case vector x has components a,b,c,...

CornMuffin, how would you know to start off with a 2x2?

Well, the identity and zero matrix are the only solutions I can think of for any A, such that AB=BA. Unless they just want some scalar multiple of the identity matrix, which I think would work (trying that now to be sure). Something like [5,0; 0,5], but that's still the identity matrix basically.

9. May 17, 2012

Steely Dan

Sure, but it's not the identity matrix, so it is one possible solution :)

10. May 17, 2012

QuarkCharmer

If that's the answer they were looking for that is pretty lame. There still "could" be some matrix B which satisfies this argument. I'll go with some cheap multiple of the identity matrix for now and revisit after I read the rest of chapter 2.

Thanks for the help

11. May 17, 2012

CornMuffin

Sometimes it is easier to start off with a simpler matrix... by using any simple 2x2 matrix first, it might help you find a matrix B with a larger matrix

edit: there are other matrices that would work that is not a multiple of the identity

12. May 17, 2012

Dustinsfl

Once you get all your equations set equal to each other, you have
$$\begin{pmatrix} 0&-1&-1&1&0&0&1&0&0&:0\\ 1&1&4&0&-1&0&0&-1&0&:0\\ -1&-3&-2&0&0&1&0&0&1&:0\\ 1&0&0&1&-1&-1&3&0&0&:0\\ 0&1&0&-1&0&-4&0&3&0&:0\\ 0&0&0&-1&-3&-3&0&0&3&:0\\ 1&0&0&4&0&0&4&-1&-1&:0\\ 0&1&0&0&4&0&-1&3&-4&:0\\ 0&0&1&0&0&4&-1&-3&0&:0 \end{pmatrix}$$

13. May 17, 2012

Steely Dan

I think that solving nine simultaneous equations is pretty "lame." Finding simple answers to problems that seem complicated is what is cool.

But I don't know what the authors were looking for. That depends on whether they're the type of author that has a sense of humor or not.

14. May 17, 2012

CornMuffin

you can use a scalar multiple of the matrix A, or of the inverse of A, or you can even use something like B=A^3+5A+2A^-1 :D

15. May 17, 2012

daveb

I haven't done the math, but perhaps it might be easier to find A-1 so you get B = A-1BA.

16. May 17, 2012

There's an extremely easy choice of B, an extremely easy choice...

17. May 17, 2012

Bohrok

If you make B the inverse of A, that would work; however I don't think that's what the author wants for an answer since inverses are covered in the next section...

18. May 17, 2012

Citan Uzuki

You all are making this much too complicated. Why not just let B=A (I'm guessing this is what you were getting at, sponsoredwalk).

19. May 17, 2012

Maybe A is a bit simpler than A⁵³⁵

20. May 18, 2012

I like Serena

Okay... let's try something.

B is not supposed to be the zero matrix or the identity.
So let's pick the "best" next thing...
Pick a matrix with all entries set to zero, except the top left one...

Yep, that does the trick!

I guess you will need to set more conditions...

Edit: Oh, and yes, B=A or B=A-1 will also do the trick. :)