Let $(X,\mathcal{A})$ be a ringed space. A complex $\mathcal{S}^{\bullet}$ of $\mathcal{A}$-modules is $\textit{perfect}$ if for any point $x\in X$, there exists an open neighborhood $U$ of $x$ and a bounded complex of finite rank free sheaves of $\mathcal{A}$-modules $\mathcal{E}^{\bullet}_{U}$ on $U$ such that the restriction $\mathcal{S}^{\bullet}|_U$ is quasi-isomorphic to $\mathcal{E}^{\bullet}_{U}$.

Moreover $\mathcal{S}^{\bullet}$ is $\textit{strictly perfect}$ if there exists a complex of finite rank free sheaves of $\mathcal{A}$-modules $\mathcal{E}^{\bullet}$ on $X$ such that $\mathcal{S}^{\bullet}$ is quasi-isomorphic to $\mathcal{E}^{\bullet}$.

Now I think the following statement is true:

Let $X$ be an affine scheme and $\mathcal{A}=\mathcal{O}_X$ be the sheaf of regular functions on $X$. Then any perfect complex on $(X,\mathcal{A})$ is actually a strictly perfect complex.

Is there any simple proof of the above statement?

Let $(X,\mathcal{A})$ be a ringed space. A complex $\mathcal{S}^{\bullet}$ of $\mathcal{A}$-modules is $\textit{perfect}$ if for any point $x\in X$, there exists an open neighborhood $U$ of $x$ and a bounded complex of finite rank free sheaves of $\mathcal{A}$-modules $\mathcal{E}^{\bullet}_{U}$ on $U$ such that the restriction $\mathcal{S}^{\bullet}|_U$ is quasi-isomorphic to $\mathcal{E}^{\bullet}_{U}$.

Moreover $\mathcal{S}^{\bullet}$ is $\textit{strictly perfect}$ if there exists a complex of finite rank locally free sheaves of $\mathcal{A}$-modules $\mathcal{E}^{\bullet}$ on $X$ such that $\mathcal{S}^{\bullet}$ is quasi-isomorphic to $\mathcal{E}^{\bullet}$.

Now I think the following statement is true:

Let $X$ be an affine scheme and $\mathcal{A}=\mathcal{O}_X$ be the sheaf of regular functions on $X$. Then any perfect complex on $(X,\mathcal{A})$ is actually a strictly perfect complex.

Is there any simple proof of the above statement?