Does $K_{12,12}$ decompose into $K_{4,4}-I$ subgraphs? 
Q: Does the complete bipartite graph $K_{12,12}$ decompose into $K_{4,4}-I$ subgraphs, where $I$ is a $1$-factor (i.e., a perfect matching)?

The obvious necessary conditions work:


*

*$K_{12,12}$ has $12^2$ edges which is divisible by $12$, the number of edges in $K_{4,4}-I$.

*Vertices in $K_{12,12}$ have degree $12$ which is divisible by $3$, the degree of the vertices in $K_{4,4}-I$.
I haven't put too much effort into solving this thus far.  It seems like it would require considerable effort to solve computationally, so I'm hoping there is a cleverer solution.
An alternative way of thinking about this problem:

Q: Does there exist a $12 \times 12$ matrix containing $12$ copies of each symbol in $\{1,2,\ldots,12\}$ such that each row and each column containing a copy of symbol $i$ contains exactly $3$ copies of $i$?

This question is a specific instance of another problem I'm thinking about.

[Edit (by Rebecca): the second version was correct, but let me tidy things up a bit.]
$$
\begin{array}{|cccccccccccc|}
\hline
5&1&1&1&9&2&2&2&5&5&9&9 \\
1&5&1&1&2&9&2&2&5&5&9&9\\
1&1&6&1&2&2&10&2&6&6&10&10\\
1&1&1&6&2&2&2&10&6&6&10&10\\
11&3&3&3&7&4&4&4&11&11&7&7\\
3&11&3&3&4&7&4&4&11&11&7&7\\
3&3&12&3&4&4&8&4&12&12&8&8\\
3&3&3&12&4&4&4&8&12&12&8&8\\
5&5&6&6&9&9&10&10&5&6&9&10\\
5&5&6&6&9&9&10&10&6&5&10&9\\
11&11&12&12&7&7&8&8&11&12&7&8\\
11&11&12&12&7&7&8&8&12&11&8&7\\
\hline
\end{array}
$$
In color:

And here's an illustration of the decomposition:

