Force distribution on corner supported plane This question has been annoying me for a while. If you have a completely ridged rectangular plate of width and height x and y that is supported on each corner (A,B,C,D) and has force (F) directly in its center then I think the force on each corner support will be F/4.

What I want to know is how to prove this? Obviously there are 4 unknowns so we require 4 equations. 
$$\sum F_z = 0$$
$$\therefore F_A+F_B+F_C+F_D=F$$
Also
$$\sum M = 0$$
Now taking the moments about point A 
\begin{equation*}
\begin{vmatrix}
        i & j & k \\
        x & 0 & 0 \\
        0 & 0 & F_B \\
        \end{vmatrix}
+
\begin{vmatrix}
        i & j & k \\
        x & y & 0 \\
        0 & 0 & F_C \\
        \end{vmatrix}
+\begin{vmatrix}
        i & j & k \\
        0 & y & 0 \\
        0 & 0 & F_D \\
        \end{vmatrix}+
\begin{vmatrix}
        i & j & k \\
        x/2 & y/2 & 0 \\
        0 & 0 & -F \\
        \end{vmatrix}=0
\end{equation*}
As the sum of moments equals 0, let i and j = 0
$$\therefore F_C+F_D=F/2$$ and $$F_B+F_C=F/2$$
Then taking the moments about another point to get 4th equation. But no matter what location I use the equations will not solve. I have been using a matrix to find $$F_A, F_B, F_C, F_D$$  See below. But when getting the det of the first matrix the answer is always equal to 0.
\begin{equation*}
\begin{vmatrix}
        1 & 1 & 1 & 1 \\
        0 & 0 & 1 & 1 \\
        0 & 1 & 1 & 0 \\
        ? & ? & ? & ? \\
        \end{vmatrix}
*
\begin{vmatrix}
        F_A \\
        F_B \\
        F_C \\
        F_D \\
        \end{vmatrix}
=\begin{vmatrix}
        1 \\
        0.5 \\
        0.5 \\
        ? \\
        \end{vmatrix}*F
\end{equation*}
Can someone please tell me what I am doing wrong? Thank you in advance.

I think I have a solution.  Considering the corner forces $A$, $B$, $C$ and $D$ you have a system of 3 equations and 4 unknowns
$$\begin{align} 
 A + B + C + D & = F \\
 \frac{y}{2} \left(C+D-A-B\right) &= 0 \\
 \frac{x}{2} \left(A+D-B-C\right) & = 0 
\end{align}$$
$$\begin{vmatrix}
   1 & 1 & 1 & 1 \\
  -\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
  \frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} A \\ B \\ C \\ D \end{vmatrix} = \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} $$
What if consider the forces as deviation from $\frac{F}{4}$ such that
$$\begin{align}
  A &  = \frac{F}{4} + U \\
  B &  = \frac{F}{4} + V \\
  C &  = \frac{F}{4} + W \\
  D &  = \frac{F}{4} + G \end{align} $$
$$ \begin{vmatrix}
   1 & 1 & 1 & 1 \\
  -\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
  \frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} \frac{F}{4} \\ \frac{F}{4} \\ \frac{F}{4} \\ \frac{F}{4} \end{vmatrix} +
\begin{vmatrix}
   1 & 1 & 1 & 1 \\
  -\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
  \frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} U \\ V \\ W \\ G \end{vmatrix} = \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} $$
$$ \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} +
\begin{vmatrix}
   1 & 1 & 1 & 1 \\
  -\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
  \frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} U \\ V \\ W \\ G \end{vmatrix} = \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} $$
$$ \begin{vmatrix}
   1 & 1 & 1 & 1 \\
  -\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
  \frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} U \\ V \\ W \\ G \end{vmatrix} = \begin{vmatrix} 0 \\ 0 \\ 0 \end{vmatrix} $$
Now a possible solution is $U=0$, $V=0$, $W=0$ and $G=0$
I think to beyond this method, you need to assume a non-rigid frame, with some simple stiffness matrix, and in the end find the limit where $k \rightarrow \infty$.
