Show that $AB-BA\ne C$ for every real $3\times 3$ matrices $A$ and $B$ and some specific $3\times3$ matrix $C$ 
Prove that there cannot exist real $3 \times 3$ matrices $A$ and $B$ such that 
$$AB-BA= \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 1 & 1\end{bmatrix}$$

I have no idea to solve it as both $A$ and $B$ are unknown.

You can employ the trace for this: For any two $3\times 3$-matrices $A$, $B$ we have ${\rm Tr}(AB) = {\rm Tr}(BA)$. Therefore, ${\rm Tr}(AB-BA) = 0$ for all $A$ and $B$. Since ${\rm Tr}(\begin{pmatrix}1 & 0& 1\\0 & 1 & 0\\0 & 1 & 1\end{pmatrix}) = 3\neq 0$, there exist no $A$, $B$ such that $AB-BA = \begin{pmatrix}1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 1 & 1\end{pmatrix}$.
