Yes, at least if we're not in characteristic 2.
Since all nodal cubics are projectively equivalent, it is enough to find one example. Trying a few symmetric $3 \times 3$ determinants soon turns up the matrix
$$
M = \left[ \begin{array}{ccc}
x & x & y \cr
x & z & 0 \cr
y & 0 & x
\end{array} \right]
$$
with determinant $x^2(z-x)-zy^2$. So the discriminant locus of
the associated net of conics is $zy^2 = x^2(z-x)$, which has a
node at $(x:y:z) = (0:0:1)$ [set $z=1$ to get the more familiar
affine model $y^2 = x^2 - x^3$ with a node at the origin].
At that point $M$ becomes
$$
\left[ \begin{array}{ccc}
0 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 0
\end{array} \right],
$$
where the conic degenerates to a double line as desired.