Explicit worked example of symmetrizing system of conservation laws Pg 4 of this book gives the Euler equations for an ideal gas in the variables $(p, \textbf{v}, S)$ read:
$$\frac{Dp}{Dt} + \gamma p \, \text{div} \textbf{v} = 0$$
$$\rho \frac{D\textbf{v}}{Dt} + \nabla p = 0$$
$$\frac{DS}{Dt} = 0$$
Here, $\frac{D}{Dt}$ is the material derivative and $\rho$ is a function of $S$ and $p$. The author of the comment in the second answer in this post discusses how to express the foregoing system of conservations laws in matrix form and have this matrix be symmetric. According the pg 9 of this book, multiplying the foregoing system by the diagonal 5 by 5 matrix
$$A = 
\begin{bmatrix}
(\gamma p )^{-1} &  & \\
 & \rho(p,S) \textbf{I}_3 & \\
 & & 1
\end{bmatrix}
$$
gives a symmetric coefficient matrix outlined in the spirit of the comments in the second answer on the post above. In particular, my understanding is that the system of conservation laws should have the form:
$$A\textbf{z}_t + A\,\text{div}\,\textbf{F}(\textbf{z}) = 0$$
where $\textbf{z} = (p, u, v, w, S)$ (the components of velocity are given by $u, v, w$ here). I'm having trouble expressing second term in component form so that I can see an explicit example of the symmetric matrix discussed in the post above.
 A: To see an explicit example it would help to write out the system
as follows. It is almost symmetric already, because it reads
$$
\begin{bmatrix} 1 &  0   & 0    & 0    & 0 \\
                0 & \rho & 0    & 0    & 0 \\
                0 &  0   & \rho & 0    & 0 \\
                0 &  0   &   0  & \rho & 0 \\
                0 &  0   &   0  & 0    & 1 \end{bmatrix}z_t
+
\begin{bmatrix} v_1 & \gamma p & 0        & 0        & 0 \\
                1   & \rho v_1 & 0        & 0        & 0 \\
                0   &  0       & \rho v_1 & 0        & 0 \\
                0   &  0       & 0        & \rho v_1 & 0 \\
                0   &  0       & 0        & 0        & v_1\end{bmatrix}z_{x_1}
 $$ $$
+
\begin{bmatrix} v_2 &  0       & \gamma p & 0      & 0 \\
                0   & \rho v_2 & 0        & 0        & 0 \\
                1   &  0       & \rho v_2 & 0        & 0 \\
                0   &  0       & 0        & \rho v_2 & 0 \\
                0   &  0       & 0        & 0        & v_2\end{bmatrix}z_{x_2}
+
\begin{bmatrix} v_3 &  0       & 0        & \gamma p & 0 \\
                0   & \rho v_3 & 0        & 0        & 0 \\
                0   &  0       & \rho v_3 & 0        & 0 \\
                1   &  0       & 0        & \rho v_3 & 0 \\
                0   &  0       & 0        & 0        & v_3\end{bmatrix}z_{x_3}
 = 0.
$$
It is only necessary to multiply the first line by
$(\gamma p)^{-1}$ to make the coefficient matrices symmetric.
