Answer: Is true sometimes.
Explanation:
I guess that here we have two matrices, A and B, that are nxn.
We can see that if those matrices can conmutate, then we can try it with some simple matrices.
![A = \left[\begin{array}{ccc}1&0\\0&-1\end{array}\right] . B = \left[\begin{array}{ccc}2&0\\1&1\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/tddjpm26vkzt483llbikkgd7kzblzyqej9.png)
Here, we would have that:
![AB = \left[\begin{array}{ccc}2&0\\-1&-1\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/kw0hedu30d1myb3wbmdmkm3ocr6fgg21ma.png)
![BA = \left[\begin{array}{ccc}2&0\\1&-1\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/3guwih84ztypuyieb0klqxax1eq3kymuuv.png)
You can see that AB and BA are different, then the statement is not always true.
But it is true sometimes, if A or B are the identiti, then I*A = A*I, in this case would be true.
It is also true if A and B are diagonal matrices, let's prove it:
![A = \left[\begin{array}{ccc}a&0\\0&b\end{array}\right] , B = \left[\begin{array}{ccc}c&0\\0&d\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/xxh7lbprg5auydmpyf61bipjqh74jwtluf.png)
![AB = \left[\begin{array}{ccc}ac&0\\0&bd\end{array}\right] = BA](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ssj34bw9rq956txddfzlgjoc299uzz9dy0.png)