Bruns and Herzog in their book Cohen-Macaulay Rings, page 120 write: "The Hilbert-Burch theorem 1.4.17 identifies perfect ideals of grade $2$ as the ideals of maximal minors of certain matrices".
For the convenience of the community I restate the Hilbert-Burch theorem as given by Bruns and Herzog:
Hilbert-Burch Theorem [1.4.17 in B&H]: Let $R$ be a Noetherian ring and $I$ an ideal with a free resolution $F_{\bullet}:0 \rightarrow R^n \stackrel{\phi}{\rightarrow} R^{n+1} \rightarrow I \rightarrow 0$. Then there exists an $R$-regular element $a$ such that $I = a I_n(\phi)$, ($I_n(\phi)$ is the $n$-th Fitting invariant of $\phi$). If $I$ is projective, then $I=(a)$, and if $\operatorname{projdim}I=1$, then $I$ is perfect of grade $2$. Conversely, if $\phi:R^n \rightarrow R^{n+1}$ is an $R$-linear map such that $\operatorname{grade} I_n(\phi) \ge 2$, then $I=I_n(\phi)$ has the free resolution $F_{\bullet}$.
My question: Now let $I$ be a perfect ideal of grade equal to $2$. Since $I$ is perfect we must have that $\operatorname{projdim} R/I = 2$. How does it follow then from the Hilbert-Burch theorem that $I$ is $I_n(\phi)$ for some $\phi$ and some $n$? How do we construct this $\phi$?
It follows from the first part of the statement when the ring is local. Consider the following minimal free resolution for $R/I$: $$ 0 \to R^m \to R^n \to R \to R/I \to 0. $$ Since rank is additive, we see that $m + 1 = n$. Then the map $\phi: R^m \to R^n$ can be written as $m \times n$ matrix. The Theorem says $I = aI_n(\phi)$ where $a$ is an regular element. If $a$ is not a unit, then height of $I$ is at most $1$. Therefore, $a$ is a unit and $I = I_n(\phi)$.