Is this possible? AB- BA=I

For a matrix $A=[a_{ij}]$ of size $n\times n$, its trace $Tr(A)$ is defined by $$ Tr(A)=\sum_{i=1}^n a_{ii} $$ . You can verify it yourself that $$ Tr(AB)=Tr(BA)$$ and that $$ Tr(A+B)=Tr(A)+Tr(B) $$

Therefore if $AB-BA = \Bbb I$, then we have $$n=Tr(\Bbb I)= Tr(AB-BA)= Tr(AB)-Tr(BA) = 0 $$ which is impossible.

Look at the trace of $AB$ and the trace of $BA$

you can see for example that

$$\mathrm{Tr}(AB) - \mathrm{Tr}(BA) =0\neq \mathrm{Tr}(\mathrm{I}_n)=n$$