I tried this question by multiplying explicitly the matrices but I think I'm not getting anything, so I think, well let's suppose false so $C(ABBA)=C$ and find a contradiction but also I'm not getting anything.

4Have you encountered the trace of matrices? – Tobias Kildetoft Jun 23 '16 at 19:41

1Hint: $tr(AB)=tr(BA)$. Assuming your ground field doesn't have characteristic two. Also, this is surely a duplicate. – Jyrki Lahtonen Jun 23 '16 at 19:41

1I vote to reopen. This is specifically asking about $2\times 2$ matrices and here brute force can be used. – Thomas Jun 23 '16 at 19:46

1See also [this general version](http://math.stackexchange.com/q/99175/11619). – Jyrki Lahtonen Jun 23 '16 at 19:48

Here is [another question about $2\times2$ matrices](http://math.stackexchange.com/questions/954959/provethatabbai2cannotholdwhateverthereal2times2matriesab). But all solutions there use trace. – Martin Sleziak Jun 24 '16 at 08:05
3 Answers
You could just brute force it: $$\begin{align} A =\pmatrix{a_1 & a_2 \\ a_3 & a_4} \quad B =\pmatrix{b_1 & b_2 \\ b_3 & b_4} \\ \end{align} $$ $$\begin{align} AB  BA &= \pmatrix{a_1 & a_2 \\ a_3 & a_4}\pmatrix{b_1 & b_2 \\ b_3 & b_4}  \pmatrix{b_1 & b_2 \\ b_3 & b_4}\pmatrix{a_1 & a_2 \\ a_3 & a_4} \\ &= \pmatrix{a_1b_1 + a_2b_3& a_1b_2 + a_2b_4 \\ ? & ?}  \pmatrix{a_1b_1 + a_3b_2& ? \\ ? & ?} \end{align} $$ Now just show that this can't happen. (Yes, there is work to do for you.)
As you mention in your question, when you have found $AB  BA$, you could (maybe) use that $C(AB  BA) = (ABBA)C$ for all $C$. For example this would have to be true for $$ C = \pmatrix{1 & 0 \\ 0 & 0} $$ and matrices like this.
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Hint: Given two matrices $A, B \in M_n(\mathbb{F})$, we always have $\operatorname{trace}(AB) = \operatorname{trace}(BA)$ as can be verified by a direct calculation (and in fact, this property characterizes the trace as a linear map uniquely up to normalization). Apply $\operatorname{trace}$ to your equation to obtain a contradiction (assuming $\operatorname{char} \mathbb{F} \neq 2$).
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The trace is the unique linear functional $\varphi \colon M_n(\mathbb{F}) \rightarrow \mathbb{F}$ that satisfies $\operatorname{trace}(AB) = \operatorname{trace}(BA)$ and $\operatorname{trace}(I_n) = n$. – levap Jun 23 '16 at 19:45

So by characterizes, you meant "as a linear map". Anyway, it seems strange to specifically put a field into this without noting the possible exceptions (which are probably not relevant for the OP, but still). – Tobias Kildetoft Jun 23 '16 at 19:46

@TobiasKildetoft Thanks for the comment, I've edited the answer to make it more precise. – levap Jun 23 '16 at 19:49
Assuming we're working over a field with characteristic $\;\neq2\;$ :
$$ABBA=I_2\implies 0=Tr.(AB)Tr.(BA)=Tr.(ABBA)=Tr.(I_2)=2$$
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