<s>[INST] Solution to a system of linear equations in GF(2) Denote the system in $GF(2)$ as $Ax=b$, where:
$$
\begin{align}
A=&(A_{ij})_{m\times m}\\
A_{ij}=&
\begin{cases}
(1)_{n\times n}&\text{if }i=j\quad\text{(a matrix where entries are all 1's)}\\
I_n&\text{if }i\ne j\quad\text{(the identity matrix)}
\end{cases}
\end{align}
$$
that is, $A$ is a square matrix of order $m\times n$. And $b$ is a 0-1 vector with length $m\times n$. Now what is the solution of this system, if any, for a general pair of $m$ and $n$?
Example: For $m=2,n=3$ and $b=(0, 1, 0, 0, 1, 0)^T$, we have
$$
A=
\begin{pmatrix}
 1 & 1 & 1 & 1 & 0 & 0 \\
 1 & 1 & 1 & 0 & 1 & 0 \\
 1 & 1 & 1 & 0 & 0 & 1 \\
 1 & 0 & 0 & 1 & 1 & 1 \\
 0 & 1 & 0 & 1 & 1 & 1 \\
 0 & 0 & 1 & 1 & 1 & 1
\end{pmatrix}
$$
then one solution is $x=(1, 0, 1, 0, 1, 0)^T$
I know Gaussian elimination. I am trying but find it not very easy when dealing with a general case.
 [/INST] Some observations, too long for a comment:
If $n$ is odd, your matrix is not invertible, and so there is no solution for arbitrary $b$ (and a solution will not be unique if it exists). First, do some row operations to rewrite the constituent blocks to $$\pmatrix{1&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\1&1&0&0\\1&0&1&0\\1&0&0&1}$$
Then do some column operations to rewrite the blocks to
$$\pmatrix{n\bmod 2&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$$
But if $n$ is odd, then columns $1$, $n+1$, $2n+1$, .. $(m-1)n+1$ are now all identical.
On the other hand, if $m$ is odd, then permuting the indices will turn the $m\times n$ problem into an $n\times n$ problem from the same family, and that will not be invertible either.
Some further progress in the even case (before I noticed Robert's elegant solution): We have the blocks
$$\pmatrix{0&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$$
Further column operations give
$$\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\0&1&1&1\\0&0&1&0\\0&0&0&1}$$
and by row operations we can blank out the new 1's to the right:
$$\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$$
Now everything decouples into one $2m\times 2m$ problem (containing the first two rows and columns of each block and $n-2$ separate $m\times m$ problems.
</s>