Classifying linear first-order PDE system (elliptic, hyperbolic, or parabolic) 

  
*Consider the constants
  $$\begin{aligned}
& (\text i.)\; a_1 = b_1 = a_2 = b_2 = 1 \\
& (\text {ii}.)\; a_1 = b_2 = 1, \quad b_1 = 0, \quad a_2 = -1 \\
& (\text {iii}.)\; a_1 = b_1 = b_2 = 1, \quad a_2 = -1
\end{aligned} \tag{3}$$
  classify the following system for the constants given above
  $$\begin{aligned}
& a_1 u_x + a_2 v_y = g_1 \\
& b_1 v_x + b_2 u_y = g_2
\end{aligned} \tag{4}$$
  where $g_1$ and $g_2$ are +ive constants.
  

I have already put the system in the form 
$$
Aq_x + Bq_y = C,
$$ 
with 
$$
A =
\begin{bmatrix} 
a_1 & 0\\ 
0 &b_1
\end{bmatrix},\quad B=
\begin{bmatrix} 
0 & a_2\\ 
b_2 & 0
\end{bmatrix},\quad q=
\begin{pmatrix} 
u\\ 
v
\end{pmatrix}.
$$
But then, I do not know how to continue from here on to find out the eigenvalues due to the term 
$$
C = 
\begin{pmatrix} 
g_1\\ 
g_2
\end{pmatrix}
$$
that is confusing me. 
If $C$ were $0$, then I'd know how to proceed, but unfortunately this is not the case. 
Any help would be appreciated.
 A: Let us differentiate the first equation w.r.t. $y$ and the second one w.r.t. $x$:
\begin{aligned}
a_1 u_{xy} + a_2 v_{yy} &= 0 \\
b_1 v_{xx} + b_2 u_{yx} &= 0
\end{aligned}
Using the equality of mixed derivatives for $u$ and the fact that $a_1 = b_2 = 1 \neq 0$, we obtain the equations
$$
u_{xy} = u_{yx} = -a_2/a_1 v_{yy} = -b_1/b_2 v_{xx}
$$
Since $a_2\neq 0$, we are left with the second-order linear PDE $A v_{xx} + 2 B v_{xy} + Cv_{yy} = 0$ satisfied by $v$, with $A = \frac{a_1 b_1}{a_2 b_2}$, $B = 0$ and $C = -1$.


*

*The case (i) is hyperbolic since $B^2 - A C > 0$.

*The case (ii) is parabolic since $B^2 - A C = 0$.

*The case (iii) is elliptic since $B^2 - A C < 0$.


An alternative derivation would consist in the diagonalization of the matrices
$$
\mathcal A = \begin{bmatrix}
a_1 & 0 \\
0 & b_1 
\end{bmatrix} \qquad\text{and}\qquad \mathcal B = \begin{bmatrix}
0 & a_2 \\
b_2 & 0 
\end{bmatrix}
$$
as $\mathcal A = P D_{\mathcal A} P^{-1}$ and $\mathcal B = P D_{\mathcal B} P^{-1}$, where $D_{\mathcal A}$, $D_{\mathcal B}$ are diagonal matrices.
For instance, (i) leads to
$$
P = \begin{bmatrix}
-1 & 1 \\
1 & 1 
\end{bmatrix} \qquad\text{and}\qquad P^{-1} = \begin{bmatrix}
-1/2 & 1/2 \\
1/2 & 1/2
\end{bmatrix}
$$
so that
$D_{\mathcal A} = \text{diag}(1,1)$ and $D_{\mathcal B} = \text{diag}(-1,1)$.
Hence, the first-order system $\mathcal A q_x + \mathcal B q_y = g$ with $q = (u,v)^\top$ and $g = (g_1, g_2)^\top$ rewrites as $p_x + D_{\mathcal B} p_y = f$ with $p = P^{-1} q$ and $f = P^{-1} g$. Since $D_{\mathcal B}$ has real eigenvalues, this system is a first-order hyperbolic system.
