= Solution
Write $u=u_x$ and $B=B_z$. In this geometry the <Lorentz force density> is purely longitudinal and equals $-\partial_x(B^2/2\mu_0)\mathbf e_x$. The <ideal magnetohydrodynamic induction equation> becomes $B_t+\partial_x(uB)=0$. Together with the <continuity equation> and adiabatic <pressure> evolution, the four primitive-variable equations are
$$
\rho_t+u\rho_x+\rho u_x=0,\quad
p_t+up_x+\gamma p u_x=0,\quad
u_t+uu_x+\rho^{-1}p_x+\frac{B}{\mu_0\rho}B_x=0,\quad
B_t+uB_x+Bu_x=0.
$$
They have the required quasilinear form $\mathbf U_t+A\mathbf U_x=0$ with
$$
A=\begin{pmatrix}
u&0&\rho&0\\0&u&\gamma p&0\\0&1/\rho&u&B/(\mu_0\rho)\\0&0&B&u
\end{pmatrix}.
$$
Here the notation $u_x$ in derivatives means $\partial_xu$, whereas $u$ denotes the longitudinal <velocity>. Spatial dependence is one-dimensional; time dependence is allowed in the wave problem.
Put $c_f^2=\gamma p/\rho+B^2/(\mu_0\rho)$, and assume $\rho>0$ and $c_f>0$. Expansion of the <characteristic polynomial> gives
$$
\det(A-\lambda I)=(u-\lambda)^2\bigl[(u-\lambda)^2-c_f^2\bigr],\qquad
\boxed{\lambda_0=u\text{ twice},\quad\lambda_\pm=u\pm c_f.}
$$
Two independent <eigenvectors> for $\lambda_0$ can be chosen as $(1,0,0,0)^T$ and $(0,-B/\mu_0,0,1)^T$. For the two propagating branches, useful <eigenvectors> are
$$
r_\pm=\left(1,\frac{\gamma p}{\rho},\frac{\pm c_f}{\rho},\frac B\rho\right)^T.
$$
The repeated branch advects an <entropy> disturbance and a disturbance with unchanged total gas-plus-magnetic <pressure>. The $\pm$ branches are perpendicular <fast magnetosonic waves>, with speed $c_f$ relative to the fluid.
On a uniform background, a Fourier amplitude $\widehat{\mathbf U}e^{i(kx-\omega t)}$ obeys $(kA-\omega I)\widehat{\mathbf U}=0$. Thus \b[the <eigenvalues> are the <phase velocities> $\omega/k$] for $k\ne0$, and the four dispersion branches are $\omega=ku$ twice and $\omega=k(u\pm c_f)$. The reduced geometry retains neither an independent transverse <velocity> nor an additional propagating <Alfvén wave>.
To construct each finite-amplitude <simple wave in magnetohydrodynamics>, trace a state-space curve with tangent $r_\pm$, using $\rho$ as its parameter. Integrating the <pressure> and magnetic components gives
$$
p=K\rho^\gamma,\qquad B=b\rho,\qquad
\frac{du}{d\rho}=\pm\frac{c_f(\rho)}{\rho},\qquad
c_f^2=\gamma K\rho^{\gamma-1}+\frac{b^2\rho}{\mu_0}.
$$
The constants $K$ and $b$ are the <entropy> and transverse <magnetic flux freezing> invariants along this state curve. Hence
$$
u(\rho)=u_*\pm\int_{\rho_*}^{\rho}\frac{c_f(r)}r\,dr.
$$
Since $A\mathbf U_\rho=\lambda_\pm\mathbf U_\rho$, substituting this curve into the original system reduces it to $\rho_t+\lambda_\pm(\rho)\rho_x=0$. By the <chain rule>,
$$
\boxed{\partial_t\lambda_\pm+\lambda_\pm\partial_x\lambda_\pm=0.}
$$
Thus the <characteristic speed> itself satisfies the <Inviscid Burgers equation>. For initial speed $L(a)$, its smooth solution is $\lambda_\pm=L(a)$ along $x=a+tL(a)$, until the <characteristic curves> intersect. This derivation supplies genuine state curves for <simple waves>, not merely a linear <dispersion relation>.
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