Converting a PDE to a matrix form I have to solve the problem
$U_{xx}-U_{xy}+2U_y+Uyy-3U_{yx}+4U=0$
using diagonal matrix as described in this article page 44 section 3.2
But my problem is there the matrix A is symmetric matrix  assuming derivatives commute. But in the given problem coefficient of $U_{xy}=-1$ and coefficient of $U_{yx}=-3$.
Then the corresponding matrix
  A=$$
        \begin{bmatrix}
        1/2 & -1 \\
          \\
        -3 & 0  \\
        \end{bmatrix}
$$
Then the matrix A is not symmetric
 A: As posted in the commentaries depends on the hypothesis about the solutions. First I'm going to assume $U_{xy}=U_{yx}$.
On your article the derivatives of grade $2$ are writed as the next form:
$$\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}\frac{\partial^2u}{\partial x_i\, \partial x_j} $$
What can be writed as:
$$\sum_{i=1}^{n}a_{ii}+\sum_{i=1}^{n}\sum_{j=1,i\neq j}^{n}a_{ij}\frac{\partial^2u}{\partial x_i\, \partial x_j} $$
Assuming that your derivatives commute:
$$\sum_{i=1}^{n}a_{ii}+\sum_{i=1}^{n}\sum_{j<i}^{n}(a_{ij}+a_{ji})\frac{\partial^2u}{\partial x_i\, \partial x_j} $$
Here is the keypoint exist more than one matrix $A$ that match with this pattern for example:
$$
  A=      \begin{bmatrix}
        1 & -1 \\
          \\
        -3 & 1  \\
        \end{bmatrix}
$$ 
But if you assume that the matrix is symmetric:
$$
  A=      \begin{bmatrix}
        1 & -2 \\
          \\
        -2 & 1  \\
        \end{bmatrix}
$$ 
What is the result of putting your equation in the next way:
$$U_{xx}-U_{xy}+2U_y+Uyy-U_{yx}-2U_{yx}+4U=0$$
$$U_{xx}-U_{xy}+2U_y+Uyy-U_{xy}-2U_{yx}+4U=0$$ 
$$U_{xx}-2U_{xy}+2U_y+Uyy-2U_{yx}+4U=0$$
Using the fact that $U_{xy}=U_{yx}$. The advantage of assuming $A$ symmetric allows you to use the diagonalization tool as you can see in rest of the article. 
