Is the Cube of matrix associative?

[TheMercury79]

Homework Statement

I was wondering the same thing as the poster of this original thread:
https://www.physicsforums.com/threads/cubing-a-matrix.451979/

I have a matrix that needs to be cubed, so which order should I use:
[A]^3 = [A]^2[A] or [A][A]^2 ?

Relevant Equations

The two people that answered both say the order doesn't matter since
matrix multiplication is associative:

(A*A)*A=A*(A*A)

But I actually don't get the same matrix. What I get is the transpose of the other when I change the order

i.e when I do [A]^2[A] I get the transpose of [A][A]^2 and vice versa

What I'm trying to do is find the cube of the expectation value of x in the harmonic oscillator in matrix form.
We're supposed to use direct matrix multiplication (using ladder operators is in part b of the problem).
I seem to have got the correct matrix elements for x^2, but for x^3 the order of the multiplication does seem
to matter.

 

[fresh_42]

Homework Statement:: I was wondering the same thing as the poster of this original thread:
https://www.physicsforums.com/threads/cubing-a-matrix.451979/

I have a matrix that needs to be cubed, so which order should I use:
[A]^3 = [A]^2[A] or [A][A]^2 ?
Relevant Equations:: The two people that answered both say the order doesn't matter since
matrix multiplication is associative:

(A*A)*A=A*(A*A)

But I actually don't get the same matrix.

Then you have made a mistake somewhere. Matrix multiplication is indeed associative and thus the order irrelevant.

 

[TheMercury79]

Then you have made a mistake somewhere. Matrix multiplication is indeed associative and thus the order irrelevant.

Thanks. It turned out they are the same. I just ended up with different expressions on the transposes.
e.g (3/2)*sqrt(1/2) was transposed with sqrt(1/2)*(1+sqrt(1/2)), but these are equal so there was no problem

 

[fresh_42]

Transposing changes the order: $(A\cdot B)^\tau=B^\tau \cdot A^\tau$, same does inversion: $(A\cdot B)^{-1}=B^{-1} \cdot A^{-1}$.