the strategy about a $0-1$ matrix game Given a $4\times4$ binary matrix as following,
$\left( \begin{matrix}
   1 & 0 & 0 & 0  \\
   0 & 0 & 0 & 0  \\
   0 & 1 & 0 & 0  \\
   0 & 0 & 1 & 0  \\
\end{matrix} \right)$
if you choose $a_{ij}$ in this matrix, then all the number in i th row and j th column will be changed( from 0 to 1 or from 1 to 0).
If all number in matrix turns to $0$, you win the game.
the steps may described like $a_{11}-a_{12}-a_{32}$.
Can any given $0-1$ matrix turn to a 0 matrix?
What is the strategy?
Thanks.
 A: This is yet another variant of the Lights Out puzzle. For each button (i,j) we define a toggle matrix $T_{ij}$ where the entry is 1 if the button in that location changes state, or if it doesn't. 
For example, $T_{11}=\left(\begin{matrix}1&1&1&1&\\1&0&0&0&\\1&0&0&0\\1&0&0&0\end{matrix}\right)$
The initial matrix 
$A=\left( \begin{matrix}
   1 & 0 & 0 & 0  \\
   0 & 0 & 0 & 0  \\
   0 & 1 & 0 & 0  \\
   0 & 0 & 1 & 0  \\
\end{matrix} \right)$
The problem turns to a solving a linear equations in {0,1} system.
$A+\sum{x_{ij}T_{ij}}=0(\mod2)$, $x_{ij}\in\{0,1\}$
which is equivalent to
$\sum{x_{ij}T_{ij}}=A(\mod2)$, $x_{ij}\in\{0,1\}$
then it becomes a linear equations as followings:
$\left(\begin{matrix}
1 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 \\
0 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\
0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\
\end{matrix}\right)(x_{11},x_{12},x_{13},\cdots,x_{33})^T=(a_{11},a_{12},a_{13},\cdots,a_{33})^T$
