Given, $A, B, C$ be $2 \times2$ matrices with entries from the set of real numbers. Define $A*B = \frac{1}{2}(AB'+A'B)$ where $X'$ denotes the transpose of $X$. Does $A*B=B*A$?
My book answers this as yes, but I cannot understand why that would be, generally.
I can see from here, that, $$A*B = \frac{1}{2}(AB'+A'B)$$ and, $$B*A = \frac{1}{2}(BA'+B'A)$$
But, my understanding is that matrices $A, B$, in general, do not commute. So, in general, $AB' \neq B'A$ and $A'B \neq BA'$
So, how would $A*B=B*A?$ Can anyone justify the book's statement, or is it a mistake and I'm correct here?