Constructing Resolution for a Determinantal Ideal For some  context, this question is being asked in the context of a summer project within my university in which we are investigating equivariant resolutions of certain modules in order to give us dimensions of graded  components of those modules (which helps in computing invariants). In the case where there are only $4$ variables (and $3$ relations in the minors of a matrix) we get that the dimension of graded components are:
$$ \text{dim}\left(\mathbb{C}[a_0, a_1, a_2, a_3]/\left(\sigma_1, \sigma_2, \sigma_3 \right) \right)= 3n+1 $$ where n is the specific grading of the ring and the $\sigma_i$ are the relations from the matrix.  Now, to generalise I have a ring of polynomials $R=\mathbb{C}[a_0, a_1, a_2, a_3, a_4]$ and an ideal generated by the $2\times 2$ minors of the matrix given by:
$$ \begin{pmatrix} a_0 & a_1 & a_2 & a_3 \\
                    a_1 & a_2 & a_3 & a_4 \end{pmatrix} $$
which are
$$ \Delta_0 = a_2a_4-a^2_3, \Delta_1 = a_1a_4-a_3a_2, \Delta_2 = a_1a_3 -a^2_2 \\
    \Delta_3 = a_0a_4-a_3a_1, \Delta_4 = a_0a_3-a_1a_2, \Delta_5 = a_0a_2-a^2_1 $$
Calling the ideal generated by these $I$, I want to construct a resolution (exact sequence) for $R/I$. I currently have part of it:
$$ \displaystyle
\overset{\varphi_3}\longrightarrow \bigoplus^6 R\overset{\varphi_2}\longrightarrow R \overset{\varphi_1 =\pi}\longrightarrow   R/I \longrightarrow 0 $$
where $\pi$ is the projection onto the quotient and $\varphi_2(a,b,c,d,e,f)=a\Delta_0+b\Delta_1+c\Delta_2+d\Delta_3+e\Delta_4+f\Delta_5$. After this point I'm a bit stumped. I'm not even sure what the module should be for the next map, never mind the actual map. I had thought it could be something to do with the matrix? Any help is appreciated.
$\textbf{EDIT:}$ I have since discovered that the next module in the exact sequence is $\displaystyle \bigoplus^8$ which means that $\varphi_3$ must be a $6\times 8$ matrix. The main question now is what explicitly is the map $\varphi_3$?
 A: This has since been solved in a much more simple (and probably more universal) way using exterior algebras. It was enlightening enough that it made sense to answer this question. Let
$$  a\left(\mathbf{x} \right) = a_0x^4+4a_1x^3y+6a_2x^2y^2+4a_3xy^3+a_4y^4 $$
The relations are, as mentioned above captured by the minors of the matrix given by:
$$  \begin{vmatrix}
    a_0 & a_1 & a_2 & a_3 \\
    a_1 & a_2 & a_3 & a_4
    \end{vmatrix}$$
Now, the idea is, instead of taking a quotient of $\mathbb{C}[a_0, \ldots, a_4]$ by the ideal generated by these minors, we should capture the relations in some wedge of one-forms. In this vain, define
$$  \Gamma_0 = a_0x^3+3a_1x^2y+3a_2xy^2+a_3y^3 \\
    \Gamma_1 = a_1x^3+3a_2x^2y+3a_3xy^2+a_4y^3$$
If we consider the two-form $\Gamma_0 \wedge \Gamma_1$, the coefficients are all scalar products of the minors of the above matrix. We can view these one-forms as sitting inside $\mathbb{C}^4$ and so the space we now want a resolution for is $\displaystyle \bigwedge^4 \mathbb{C}^4/\left(\Gamma_0 \wedge \Gamma_1 \right) $. The following is an exact sequence:
$$ \displaystyle0 \longrightarrow \bigoplus^3 \mathbb{C} \overset{\varphi_3} \longrightarrow \bigoplus^2 \bigwedge^1 \mathbb{C}^4 \overset{\varphi_2}\longrightarrow \bigwedge^2 \mathbb{C}^4 \overset{\varphi_1} \longrightarrow \bigwedge^4 \mathbb{C}^4 \overset{\pi} \longrightarrow \bigwedge^4 \mathbb{C}^4/\left(\Gamma_0 \wedge \Gamma_1 \right) \longrightarrow 0$$
where $\pi$ projects onto the quotient, $\varphi_1$ takes a two-form $\gamma$ and maps it to $\gamma \mapsto \gamma\wedge\left(\Gamma_0\wedge\Gamma_1 \right)$, $\varphi_2\left(\alpha, \beta \right)=\alpha\wedge\Gamma_0+\beta\wedge\Gamma_1$ and the function $\varphi_3$ takes triples of scalars $\left(\lambda, \mu, \sigma\right)\mapsto \left(\lambda\Gamma_0+\mu\Gamma_1, \mu\Gamma_1+\sigma\Gamma_2 \right)$ which is an injection. Now, importantly we want to grade the sequence and compute the dimensions of the graded parts of the quotient space. We grade as follows:
$$ \displaystyle  0 \longrightarrow \left(\bigoplus^3 \mathbb{C} \right)^{(n)} \overset{\varphi_3} \longrightarrow \left(\bigoplus^2\bigwedge^1 \mathbb{C}^4 \right)^{(n+1)} \overset{\varphi_2} \longrightarrow \left(\bigwedge^2 \mathbb{C}^4 \right)^{(n+2)} \overset{\varphi_1}\longrightarrow \left(\bigwedge^4 \mathbb{C}^4 \right)^{(n+4)} \\ \overset{\pi} \longrightarrow \left(\bigwedge^4 \mathbb{C}^4/\left(\Gamma_0\wedge \Gamma_1 \right) \right)^{(n+4)} \longrightarrow 0$$
Let $D_{n+4}=\text{dim}\left(\bigwedge^4 \mathbb{C}^4/\left(\Gamma_0\wedge \Gamma_1 \right) \right)^{(n+4)}$. We can use the dimensionality relations for an exact sequence to obtain that:
$$D_{n+4} =\frac{1}{24}\left(\binom{n+8}{4}-6\binom{n+6}{4}+8\binom{n+5}{4}-3\binom{n+4}{4}\right) \\
    = \frac{1}{24}\left(96n+408\right) \\
    = 4n+17 $$
Then, rewriting in standard indexing we get that:
$$D_n = 4n+1 $$
as we would expect.
