Construction of a connection on a vector bundle $Eto M$ with trivial determinant bundle

I read the following statement, and I am trying to see why it is true.

Let $Eto M$ be a rank 2 complex vector bundle over a smooth manifold $M$, such that the determinant bundle $det(E)=Lambda^2(E)$ is the trivial line bundle over $M$, with trivialization $phi:det(E)to Mtimes mathbb{C}$. Then there exists a connection $nabla^E$ on $E$ which induces the trivial connection on $det(E)$, under $phi$.

I wanted to construct $nabla^E$ by defining local connections, which I then glue together by a partition of unity. Let ${U_alpha}$ be a trivializing open cover of $M$. Over each $U_alpha$, I have a local frame ${e_1^alpha,e_2^alpha}$ such that $e_1^alphawedge e_2^alpha$ corresponds to 1 under the diffeomorphism $phi$. On each $U_alpha$, I can define a flat connection $nabla^alpha$ (so by declaring the sections $e^alpha_i$ to be flat). If $(rho_alpha)$ is a partition of unity subordinated to the trivializing open cover, I define $nabla^E(s)=sumlimits_alpha nabla^alpha(rho_alpha s)$, which is indeed a connection. Next, I want to check that it induces the trivial connection on $det(E)$ under $phi$. Note that $phi^{-1}(1)=sumlimits_alpha rho_alpha e_1^alphawedge e_2^alpha.$ Then
begin{align*}
nabla^{det(E)}big(sum_alpha rho_alpha e_1^alphawedge e_2^alphabig)&=sum_alpha nabla^{det(E)}big(rho_alpha e_1^alphawedge e_2^alpha)\&=sum_alpha nabla^E(rho_alpha e^alpha_1)wedge e_2^alpha+rho_alpha e_1^alphawedge nabla^E(e_2^alpha)\&=sum_alpha rho_alphabig(nabla^E(e_1^alpha)wedge e_2^alpha+e_1^alphawedge nabla^E(e_2^alpha)big).
end{align*}
I want this to be zero, but the problem I encounter is that $nabla^E(e_i^alpha)neqnabla^alpha(e_i^alpha)=0$. Nevertheless, is this the correct approach to take and how can I fix it? Does summing over all $alpha$ make sure it is correct again?

EDIT: This question seems to have been partially asked here: Connections on Bundles with Trivial Determinant
But it seems that an answer is missing and I am very curious, it seems to take a different approach, by introducing a Hermitean metric on $E$.