Prove that there are not two matrices 2x2 such that: $AB-BA=I_2$ 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 $C(AB-BA)=C$ and find a contradiction but also I'm not getting anything.
 A: 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 $C(AB - BA) = (AB-BA)C$ for all $C$. For example this would have to be true for 
$$
C = \pmatrix{1 & 0 \\ 0 & 0}
$$
and matrices like this.
A: 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 $\operatorname{char} \mathbb{F} \neq 2$).
A: Assuming we're working over a field with characteristic $\;\neq2\;$ :
$$AB-BA=I_2\implies 0=Tr.(AB)-Tr.(BA)=Tr.(AB-BA)=Tr.(I_2)=2$$
