Is the Distributive Law for Matrices Always True?

[pyroknife]

Homework Statement

Not really a homework question. Something that I've been wondering about.

The distributive law holds for matrices. Let A and B be n x n matrices.
Why is the following true for all A&B?

$(A+B)^2=A^2+2AB+B^2$

I don't undrestand that middle term (2AB) and why there's a factor of 2 there, since matrices aren't always commutative (i.e., AB doesn't always equal BA).

Shouldn't it be $(A+B)^2=A^2+AB+BA+B^2$ instead?

Homework Equations

The Attempt at a Solution

 

[robphy]

As you say, AB doesn't always equal BA.
So, as you say, $(A+B)^2=A^2+AB+BA+B^2$

But what you wrote isn't really the distributive law.
$C(A+B)=CA+CB$

 

[pyroknife]

As you say, AB doesn't always equal BA.
So, as you say, $(A+B)^2=A^2+AB+BA+B^2$

Yes, so it should be what you just stated right and not the first expression I had?

 

[Astronuc]

$(A+B)^2=A^2+AB+BA+B^2$ is correct.

Only special cases can be $(A+B)^2=A^2+2AB+B^2$. Only if AB = BA
See slide 10 of http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect1-06web.pdf

Where did one find the former equation?

 

[pyroknife]

$(A+B)^2=A^2+AB+BA+B^2$ is correct.

Only special cases can be $(A+B)^2=A^2+2AB+B^2$. Only if AB = BA
See slide 10 of http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect1-06web.pdf

Where did one find the former equation?

It was from an example from a course I took. I did miss something however, the example had also stated that BA=0 in the problem statement...but that would still not yield the former equation. I do make mistakes sometimes when I copy down notes, so maybe I wrote the 2 in their by accident or my professor did...that would be my only explanation as the answer would be $A^2+AB+B^2$ if BA=0.