Symmetric matrix with given determinant

The matrix \begin{equation} A := \begin{pmatrix} x & 0 & 0 & z \\ 0 & y & 0 & x \\ 0 & 0 & z & y \\ y & z & x & w \end{pmatrix} \end{equation} has determinant \begin{equation} \det A = -x^2y^2 - x^2z^2 -y^2z^2+ xyzw, \end{equation} which describes the Steiner surface. However, $A$ is not symmetric, so it does not answer whether the Steiner surface is a symmetroid or not.

Is it possible to find a symmetric matrix $S$ such that each nonzero entry in $S$ is a linear form in $x, y, z, w$ and $\det S = \lambda \det A$ for some $\lambda \in \mathbb{R} \setminus \{0\}$?

This is possible with $\lambda=-1$; as verified by wolfram alpha, $$ \begin{vmatrix} -x&z&0&0\\ z&x&w/2&z\\ 0&w/2&x&y\\ 0&z&y&0 \end{vmatrix}=x^2y^2+y^2z^2+z^2x^2-xyzw. $$ However it is not possible (using real coefficients) with $\lambda=1$, as follows:

Let $\mathrm{Sym}_n$ denote the set of real $n\times n$ symmetric matrices, and let $U(p,q)$ denote the set of $A\in\mathrm{Sym}_n$ with $p$ positive and $q$ negative eigenvalues (ie with signature $(p,q,n-p-q)$). Suppose $p+q=n$. Suppose $A\in U(p,q)$. Being symmetric, $A$ has an orthonormal set of eigenvectors $v_1,\ldots,v_n$ with eigenvalues $\lambda_1,\ldots,\lambda_n$. Given $\mu_1,\ldots,\mu_n\in\mathbb R\setminus\{0\}$, there is a symmetric matrix $D$ such that $Dv_i=\mu_iv_i$, so $D^TAD$ has eigenvalues $\mu_1^2\lambda_1,\ldots,\mu_n^2\lambda_n$. Therefore we can choose $D$ so that these eigenvalues are distinct. By Sylvester's law of inertia, $D^TU(p,q)D=U(p,q)$.

We will show $A$ has an open neighborhood (in the Euclidean topology) contained in $U(p,q)$. By the above we may suppose $A$ has distinct eigenvalues. Let $p(\lambda)=\det(\lambda I-A)$ be the characteristic polynomial of $A$. We can choose real numbers $\nu_0<\nu_1<\ldots<\nu_n$ between the eigenvalues of $A$ so that $$ p(\nu_n),p(\nu_{n-2}),\ldots>0, $$ $$ p(\nu_{n-1}),p(\nu_{n-3}),\ldots<0. $$ Moreover we may take $\nu_p=0$. Note that these equation ensure $A\in U(p,q)$, and they hold on an open neighborhood of $A$, as required. Therefore $U(p,q)$ is open set.

Now suppose there is a symmetric matrix $S$ linear in $x,y,z,w$ with $$ \det(S)=-x^2y^2-y^2z^2-z^2x^2+xyzw $$ (necessarily $4\times4$ from the degree). Evaluate at $w=0$ to obtain $S_0$ with $\det(S_0)=-x^2y^2-y^2z^2-z^2x^2$. Evaluating $S_0$ at values of $x,y,z$, we obtain a function $f:\mathbb R^3\rightarrow\mathrm{Sym}_4$. Let $L$ denote the set of points $(x,y,z)\in\mathbb R^3$ with at most one nonzero coordinate (ie the union of the axes). Note if $(x,y,z)\notin L$, then $\det(f(x,y,z))<0$, so $f(x,y,z)\in U(3,1)\cup U(1,3)$. However $\mathbb R^3\setminus L$ is connected, so $f(\mathbb R^3\setminus L)$ is either contained in $U(1,3)$ or $U(3,1)$. This is impossible since $U(1,3)=-U(3,1)$ and $$ f(-1,-1,-1)=-f(1,1,1). $$