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. 
$$\begin{bmatrix} &  & & \\ &  &  & \\ &  &  &\\& & & & \end{bmatrix}
\cdot\begin{bmatrix}2 & 0 &-1 & -2&-4&-4\\-1 & 1 & 0 & -1&2&-1\\ -4& 0 & 1 & 3&8&5\\ -1&0&0&1&2&1\end{bmatrix}=\begin{bmatrix}1 & 1 &3 & -1&0&2\\0 & 1 & 1 & 0&0&0\\0 & 0 & 0&-1&0& 2\\ 0&0&0&0&1&0\end{bmatrix}$$

There is a unique solution to this problem. The problem states
that $AB=C$ holds, but that someone has changed the order of the
columns in $C$ and erased what was in $A$. We are told what the matrix
$B$ is, but not told what $C$ is. We are only told what $C$ looks like
after its columns have been permuted. That is, we are told what
$CP$ is for some permutation matrix $P$ which has not been specified.
The reduced row echelon matrix for $B$ (and $C$)
plays a role, and I will label it $R$. I will write
$L$ and $M$ for the $4\times 4$-matrices for which
$LB=R=MC$. (These are uniquely determined, since $B$ and $C$ have rank 4.)
It can be shown that the matrices are:
$$
L = \left[\begin{array}{rrrr}
-1&0&-1&1\\
-2&1&-2&3\\
-1&0&0&-2\\
-1&0&-1&2
\end{array}\right],\quad
B = \left[\begin{array}{rrrrrr}
2&0&-1&-2&-4&-4\\
-1&1&0&-1&2&-1\\
-4&0&1&3&8&5\\
-1&0&0&1&2&1
\end{array}\right],
$$
$$
M = \left[\begin{array}{rrrr}
0&0&-1&0\\
0&0&0&1\\
1&-1&-1&0\\
0&1&0&0
\end{array}\right],\quad
C = \left[\begin{array}{rrrrrr}
-1&0&1&1&2&3\\
0&0&0&1&0&1\\
-1&0&0&0&2&0\\
0&1&0&0&0&0
\end{array}\right],
$$
$$
R = LB = MC
= \left[\begin{array}{rrrrrr}
1&0&0&0&-2&0\\
0&1&0&0&0&0\\
0&0&1&0&0&2\\
0&0&0&1&0&1
\end{array}\right],
$$
and
$$
A = \left[\begin{array}{rrrr}
-1&0&0&-1\\
-1&0&-1&2\\
1&0&1&-1\\
-2&1&-2&3
\end{array}\right],\quad
P = \left[\begin{array}{rrrrrr}
0&0&0&1&0&0\\
0&0&0&0&1&0\\
1&0&0&0&0&0\\
0&1&0&0&0&0\\
0&0&0&0&0&1\\
0&0&1&0&0&0
\end{array}\right].
$$

Justification:
We are told that $AB=C$. We are given $B$, but not $A$. We
are given a matrix obtained by $C$ by permuting the columns of $C$,
that is we are given $CP$ for some $6\times 6$ permutation matrix,
but we are not told what $P$ is.
Write $\textrm{Null}(X)$ for the nullspace of matrix $X$.
Since $AB=C$, it follows that $ABP=CP$, so
$x\in\textrm{Null}(CP)=\textrm{Null}(ABP)$
iff $P^{-1}x\in\textrm{Null}(AB)$.
But $B$ and $C$ have rank $4$, so $A$ must be invertible, so
$\textrm{Null}(AB)=\textrm{Null}(B)$.
To summarize, $x\in\textrm{Null}(CP)$ iff 
$P^{-1}x\in \textrm{Null}(B)$. Equivalently,
$P\cdot \textrm{Null}(B)=\textrm{Null}(CP)$.
Now we are given the matrices
$B$ and $CP$, so we can compute these nullspaces and
try to determine which permutation matrices satisfy
$P\cdot \textrm{Null}(B) =
\textrm{Null}(CP)$. 
There is only one such matrix, and it is the one
I have written above. Here I only sketch the kind of reasoning I used
to determine $P$.
Vectors $x\in \textrm{Null}(B)$ must satisfy
the following conditions.


*

*The second coordinate of $x$ must be $0$.

*The third coordinate of $x$ must be $+2$ times
the fourth coordinate of $x$ and $-2$ times the sixth coordinate of $x$.

*The first coordinate of $x$ must be $+2$ times the fifth
coordinate, but does not have to be $-2$ times another coordinate.
Now vectors in $\textrm{Null}(CP)$ must satisfy
the same conditions with different coordinate orders,
and comparing the coordinate orders we can determine $P$.
Once $P$ is known, we are basically done. We were given
$CP$, so with $P$ we can compute $C$.
From $B$ and $C$ we can solve for $A$. 
I solved for $A$
by row reducing $[B|I]$ and $[C|I]$ to $[R|L]$ and $[R|M]$ respectively,
thereby finding matrices
$L$ and $M$ such that $LB=R=MC$. Then $(M^{-1}L)B=C$, forcing
$A=M^{-1}L$. This I computed by row reducing $[M|L]$ to $[I|M^{-1}L]$.
I then used Maple to check that $AB$ is indeed a matrix that differs from $CP$ by a permutation of columns.
