Reordering of matrix multiplication

Question

I want to compute the matrix multiplication $ABA$, where $A$ and $B$ are real and orthogonal matrices. In fact, they are specifically $3\times3$ rotation matrices. However, it is much easier if I can reverse the order of $BA$ somehow, because I can then perform the multiplication much easier.

I know that matrix multiplication is not commutative, however, I am asking because both $A$ and $B$ are orthogonal matrices, and hopefully, may be there is some trick to utilize their orthogonality to reorder the product.

I tried to solve this, but got stuck here:

$$ ABA = A((BA)^{-1})^{-1}=A(A^{-1}B^{-1})^{-1} $$

Is there a way to proceed from here?

Edit: I also know that matrix multiplication is associative; however, I am not after associativity here. I want to multiply the $A$ matrix by $A$ (or by its inverse/transpose), then multiply the result by $B$.

Edit: To put this into context, consider the following product of rotation matrices $$ R_x(\theta)R_z(\pi)R_x(\theta) $$ where, $R_x(\theta)$ is the rotation matrix about the $x$-axis by $\theta$, and $R_z(\pi)$ is the rotation matrix about the $z$-axis by $\pi$.

This product simplifies to $R_z(\pi)$. Is it possible to come to this conclusion without carrying out the matrix multiplication of the three rotation matrices? It looks that the $R_x(\theta)$ got cancelled somehow.

Answer 1

Sadly, no. As @TobyMak points out, the associativity $$ A(BA) = (AB)A $$ means you can choose to evaluate either $AB$ or $BA$ but there's no trick that will allow you to work with $A^2$.

You can probably show that with an example in which $A^2 = I$.

Answer 2

$$ABA = (AB)A$$

since matrix multiplication is associative. One proof of this is on this other post on Math SE.

Answer 3

I take it that what you're trying to get at is that you have some matrix $A$ that you will use several times with several different other matrices in the place of $B.$ You would like some way to precompute a matrix $M_A$ by doing some operations on $A$ so that for any matrix $B$ you have $ABA = M_A B.$

First of all, if this is true in general it is true when $B=I,$ and therefore $AIA = M_AI$ implies $M_A = A^2.$

But then for the formula to work in general you need $ABA = A^2 B$, which implies $BA = AB,$ that is, the matrices have to commute.

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In your particular example, the rotation $R_z(\pi)$ maps the axis of $R_x(\theta)$ onto itself, but reversed. Hence the initial rotation by $\theta$ becomes a rotation by $-\theta$. You will not get such a nice result for any $B$ that is not a rotation by a multiple of $\pi$, and even when $B = R_z(\pi)$ you will not get such a nice result for any non-trivial rotation around an axis that is not either the $z$ axis or an axis in the $x,y$ plane.