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.
 A: 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.
A: An empirical formula: it looks to me like $\det(A) = (-1)^{(m-1)(n-1)} (n+m-1)(n-1)^{m-1}(m-1)^{n-1}$.  In particular, $\det(A) \equiv 0 \mod 2$ when $m$ or $n$ is odd (except in the case $m=n=1$) so, as Henning found, $A$ will not be invertible over GF(2) in those cases.
However, it will be invertible when $m$ and $n$ are both even.
EDIT: the situation over $GF(2)$ for even $m$ and $n$ is simpler than I thought.  Consider $A^2$. Let $U$ be the $n \times n$ matrix of all 1's.  Note that $U^2 = n U$.  The $(i,i)$ block of $A^2$ is $U^2 + (m-1) I = n U + (m-1) I$.  The $(i,j)$ block for $i \ne j$ is $2 U + (m-2) I$.  If $m$ and $n$ are even, the $(i,i)$ block mod 2 is $I$ and the $(i,j)$ block mod 2 is $0$ (i.e. the inverse of $A$ over $GF(2)$ is $A$). 
A: Edit: I posted a wrong answer earlier. Hope I can get things fixed this time.
For convenience, write $x^T = (x_1^T, \ldots, x_m^T)$ where each $x_i$ is a vector of length $n$. Similarly, write $b^T = (b_1^T, \ldots, b_m^T)$. Let $J$ and $u$ be respectively the $n$-by-$n$ matrix and $n$-vector with all entries equal to 1 and let $Jb_i=\beta_iu$. Your system of equations $Ax=b$ is equivalent to
$$
(\dagger): (J-I)x_i+\sum_{j=1}^m x_j= b_i\quad \forall i.
$$
Let $\ Jx_i=\alpha_iu$ where $\alpha_i\in GF(2)$ is the parity of $x_i$. So $(\dagger)$ gives
$$
(*): x_i = \sum x_j + b_i + \alpha_iu\quad \forall i.
$$
Case 1: $m$ is even. By summing up $(*)$ from $i=1,2,\ldots,m$, we get
$$\sum x_j = \sum b_j + \sum\alpha_ju.$$
Substitute this back into $(*)$, we see that the general solution to $(\ast)$ is of the form
$$
(**): x_i = \sum b_j + \sum\alpha_ju + b_i + \alpha_iu.
$$
Such $\{x_i\}$ form a solution of $(\dagger)$ if and only if $Jx_i=\alpha_iu$ for all $i$, which means
$$
\sum \beta_j + n\sum\alpha_j + \beta_i + n\alpha_i = \alpha_i
$$
or equivalently,
$$
n\sum\alpha_j  + (n-1)\alpha_i = \sum \beta_j + \beta_i.
$$
When $n$ is even, the above system has a unique solution $\alpha_i = \sum \beta_j + \beta_i$.
When $n$ is odd, the above system reduces to $\sum\alpha_j = \sum \beta_j + \beta_i$. Hence solution exists if and only if $\beta_1=\ldots=\beta_m=\beta$ and the solutions are given by $(**)$ with $\sum\alpha_j = \beta$.
Case 2: $m$ is odd. By summing up $(*)$ from $i=1,2,\ldots,m$, we get
$$\sum b_j = \sum\alpha_ju.$$
Thus a necessary condition for a solution to exist is that $\sum b_j$ is a multiple of $u$ (say, $\sum b_j=\lambda u$). If this is the case, then the general solution to $(\ast)$ is given by $x_i = y + b_i + \alpha_iu$ where $\sum \alpha_j=\lambda$ and $y$ is any $n$-vector. Such $\{x_i\}$ is a feasible solution to $(\dagger)$ if and only if $Jx_i=\alpha_iu$ for all $i$, that is, iff $Jy + \beta_iu + n\alpha_iu=\alpha_iu$. Therefore, we need $(n-1)\alpha_i+\beta_i$ to be constant and equal to the parity of $y$.
So, when $n$ is odd, solution exists only if $\beta_1=\ldots=\beta_m$. Since
$$
\sum b_j=\lambda u\ \Rightarrow\ \sum Jb_j=\lambda Ju\ \Rightarrow\ \sum\beta_j=\lambda,
$$
the previous requirement that $\sum \alpha_j=\lambda$ can be rewritten as $\sum \alpha_j=\sum \beta_j$.
When $n$ is even, that $(n-1)\alpha_i+\beta_i$ is constant means $\alpha_i+\beta_i=c$ for some constant $c$. Recall that we need $\sum \alpha_ju=\lambda u=\sum b_j$. Multiply both sides by $J$, we get $0=\sum \beta_ju$, i.e. $\sum \beta_j=0$. This is another necessary condition for the existence of solution. Suppose this is also satisfied. To make $\sum \alpha_ju=\lambda u$, we may take $\alpha_i=\beta_i$ for all $i$ if $\lambda=0$, or $\alpha_i=1-\beta_i$ for all $i$ if $\lambda=1$.
