Lights Out variant: what matrix to use when whole row+column gets flipped? I've done my research and found a few similar questions here and on other sites, but none of them have what I'm looking for.
Lights out is a simple game that has interesting math-based solutions (info here). Using the typical rules in a 5x5 board, clicking anywhere in the grid causes the clicked spot and its 4 immediate neighbours to get flipped.  But I want to implement a solution where the entire row and column get flipped. For example, in this board
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0

Clicking on the point at coordinate (4, 4) will result in
0 0 0 1 0
0 0 0 1 0
0 0 0 1 0
1 1 1 1 1 
0 0 0 1 0

Given a 5x5 board, I know how to solve it using Gaussian elimination when using the standard rules.  I read this short pdf that explained the linear algebra theory very well and showed how a solution can be obtained by solving Ax = b, where b is the column-vector of the current board and A is the following 25x25 matrix
C I 0 0 0    where    I = identity(5) and C = 1 1 0 0 0
I C I 0 0                                     1 1 1 0 0
0 I C I 0                                     0 1 1 1 0 
0 0 I C I                                     0 0 1 1 1
0 0 0 I C                                     0 0 0 1 1

I figured out how to do Gaussian elimination in modulus 2 in order to solve the matrix equation and was able to solve 5x5 boards. I also found out that by just changing the C matrix slightly, I can easily use the same code to solve 3x3.
I'm pretty sure that in order to solve the variant that I want (entire row+column gets flipped), all I need to do is figure out the C matrix.  But I tried sitting down with pencil and paper for a while and do something similar to what the PDF I linked to was doing, but I couldn't figure it out.
Does anyone know how to derive the matrix required for this?
 A: I think the matrix for your variant is this $25\times 25$ block matrix
$$A=\left(\begin{array}{ccccc}J&I&I&I&I\\ I&J&I&I&I\\I&I&J&I&I\\I&I&I&J&I\\I&I&I&I&J
\end{array}\right)$$
Where $I$ is the $5\times 5$ identity matrix and $J$ is the $5\times 5$ matrix of all $1$s.
For instance the $(1,1)$ button produces a 'toggle vector' 
$$(1 1 1 1 1 \; 1 0 0 0 0 \;1 0 0 0 0 \; 1 0 0 0 0\; 1 0 0 0 0)^T$$
and the $(1,2)$ button produces a toggle vector 
$$  (1 1111 \; 01000 \;01000 \; 01000\; 01000)^T$$
Continuing, and putting these together, give the above matrix of $I$s and $J$s.
A: Try some rows out, you will see that you have the same format. 
For example,
$$x_{1,1}+x_{1,2}+x_{1,3}+x_{1,4}+x_{1,5}+x_{2,1}+x_{3,1}+x_{4,1}+x_{5,1}=b_{1,1}$$
This gives $1$'s in the first row at the position $1,2,3,4,5,6,11,16,21$. The other rows works the same way.
The only difference in the resulting matrix is that the new $C$ is
$$\begin{array}&1&1&1&1&1\\
1&1&1&1&1\\
1&1&1&1&1\\
1&1&1&1&1\\
1&1&1&1&1\\\end{array}$$
The whole $25\times 25$ matrix is still as before. The same $I$ works.
