Find matrix $A$. How many solutions are there? The matrices below is matrix multiplcation of $AB=C$ but someone has changed the order of the coloumns in $C$ and erased what was in $A$. What is matrix $A$? Is it the only one? I have no idea how to start. I appreciate all the help I can get. 
$$\begin{bmatrix} &  & & \\ &  &  & \\ &  &  &\\& & & & \end{bmatrix}
\cdot\begin{bmatrix}2 & 0 &-1 & -2&-4&-4\\-1 & 1 & 0 & -1&2&-1\\ -4& 0 & 1 & 3&8&5\\ -1&0&0&1&2&1\end{bmatrix}=\begin{bmatrix}1 & 1 &3 & -1&0&2\\0 & 1 & 1 & 0&0&0\\0 & 0 & 0&-1&0& 2\\ 0&0&0&0&1&0\end{bmatrix}$$

Preliminary: I assume it is clear that equation $AB=C'$ (where $C'$ is a column-shuffled version of $C$) is equivalent to 7 relationships between the 7 columns of $B$ and the corresponding 7 columns of $C'$:
$$A.\begin{pmatrix}2 \\-1\\ -4\\ -1\end{pmatrix}=\begin{pmatrix} . \\ .\\ . \\ . \end{pmatrix}, \ \ A.\begin{pmatrix}0 \\1\\ 0\\ 0\end{pmatrix}=\begin{pmatrix} . \\ .\\ . \\ . \end{pmatrix}, \ \ etc...$$
Here is a solution:
$$A=\pmatrix{-1& 0& 0& -1\\-1& 0& -1& 2\\1& 0& 1& -1\\-2& 
    1& -2& 3},$$
which is such that
$$\tag{1}\pmatrix{-1& 0& 0& -1\\-1& 0& -1& 2\\1& 0& 1& -1\\-2& 
    1& -2& 3}\cdot\begin{pmatrix}2 & 0 &-1 & -2&-4&-4\\-1 & 1 & 0 & -1&2&-1\\ -4& 0 & 1 & 3&8&5\\ -1&0&0&1&2&1\end{pmatrix}=\begin{pmatrix}-1 & 0 &1 & 1&2&3\\0 & 0 & 0 & 1& 0 & 1\\ -1& 0 & 0 & 0&2&0\\ 0&1&0&0&0&0\end{pmatrix}$$
One sees that the last matrix is matrix $C$ with its columns shuffled.
The way I have obtained matrix $A$ is in 3 steps:


*

*a) by observing first that, in matrix $B$ as well as in matrix $C$, there are 2 column vectors (say $U$ and $-2U$ in $B$, and $V$ and $-2V$ in $C$) that are proportional with the same proportionality coefficient, namely columns 1 and 5 in $B$, and columns $4$ and $6$ in $C$. They have to be mapped the ones onto the others. 

*b) by reducing, in a second step, matrices $B$ and $C$ to $4 \times 4$ matrices denoted $B_1$, resp. $C_1$ by removing, in each one, the 2 columns that have been found in a). I was looking for linear combinations of columns of $B_1$ and $C_1$ giving $0$ (because neither $B_1$ nor $C_1$ is full rank). As I didn't find something significant, using a CAS, I obtained the kernels (nullspaces) of $B_1$ and $C_1$ : their bases are strikingly similar: $(0,-2,-1,1)$ for $B_1$ and
$(-2,-1,1,0)$ for $C_1$; this provides the coefficients of null linear combinations of columns I hadn't found by hand computation. This was an "invitation" to place in correspondence columns numbered 2,3,4 of $B_1$ with columns 1,2,3 of $C_1$.

*c) finally, by creating 2 new $4 \times 4$ matrices $B_2$ and $C_2$ through the following operations : i) removal in $B_1$ and in $C_1$ of one of the columns that has been used by the linear combination $-2,-1,1$, for example column 2 for $B_1$ and column 1 for $C_1$ (these columns being "charged" of the non-invertibility of $B_1$ and $C_1$ !), then  ii) "augmenting" $B_2$ and $C_2$  by "concatenation" of $U$ (as defined in part a)) as fourth column  for $B_2$ and $V$ for $C_2$. As we must have $A.B_2=C_2$, matrix $A$ is obtained by computing:
$$A=C_2.B_2^{-1}$$ 
($B_2$ is checked invertible).
As we haven't worked by necessary and sufficient conditions, we have to check (this has been done in (1)) that product $A.B$ gives a shuffled version of  $C$. 
