Let E be a vector bundle of rank 2 over a variety X. Is there a counterexample so that $E$ is not isomorphic to $E^\*\otimes det$ $E$?



counterexample related to vector bundles
As Jason Starr wrote in his comment, every rank $2$ vector bundle $E$ is isomorphic to $E^\* \otimes \det E$. Let me give a proof of this fact. 


Assume that $E$ is defined by the transition functions $$g \_{\alpha\beta} \colon U\_{\alpha} \cap U\_{\beta} \longrightarrow \mathrm{GL}(2, k), \quad g\_{\alpha \beta} = \begin{pmatrix} g\_{\alpha \beta}^{11} & g\_{\alpha \beta}^{12} \\ g\_{\alpha \beta}^{21} & g\_{\alpha \beta}^{22} \end{pmatrix}.$$
Then $E^\* \otimes \det E$ is defined by the transition functions $$h\_{\alpha \beta} = {}^tg\_{\alpha \beta}^{-1} \cdot \det g\_{\alpha \beta}= \begin{pmatrix} \,\,\,\,\,g\_{\alpha \beta}^{22} & -g\_{\alpha \beta}^{21} \\ -g\_{\alpha \beta}^{12} & \,\,\,g\_{\alpha \beta}^{11} \end{pmatrix}.$$
Now consider the family of constant functions $$f\_{\alpha} \colon U\_{\alpha} \longrightarrow \mathrm{GL}(2, k), \quad f\_{\alpha} = \begin{pmatrix} \,\,\,\,0 & 1 \\ -1 & 0 \end{pmatrix}.$$


It is immediate to check that $$h\_{\alpha \beta} = f\_{\alpha} g\_{\alpha \beta} f\_{\beta}^{-1},$$
so $\{g\_{\alpha \beta}\}$ and $\{h\_{\alpha \beta} \}$ are equivalent sets of transition functions, and this in turn means that they define isomorphic vector bundles.


