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

No solution (In case you don't shuffle $C$)
P.S: Thanks @JeanMarie
Denote the $(i,j)^{th}$ entry of $A$ by $a_{ij}$.
$$\begin{bmatrix} a_{11}  &a_{12} &a_{13} &a_{14} \\ a_{21} &a_{22}  &a_{23} &a_{24}  \\ a_{31}  &a_{33} &a_{33} &a_{34} \\a_{41} &a_{42}  &a_{43} &a_{44}  \end{bmatrix}
\cdot\begin{bmatrix}2 & 0 &-1 & -2&-4&-4\\-1 & 1 & 0 & -1&2&-1\\ -4& 0 & 1 & 3&8&5\\ -1&0&0&1&2&1\end{bmatrix}=\begin{bmatrix}1 & 1 &3 & -1&0&2\\0 & 1 & 1 & 0&0&0\\0 & 0 & 0&-1&0& 2\\ 0&0&0&0&1&0\end{bmatrix}$$ 
Multiply the first row of $A$ by the first column of $B$ (inner product), you get
$$T = 2a_{11} - a_{21} - 4a_{31} - a_{41} = 1$$
Multiply the first row of $A$ by the fifth column of $B$ (inner product), you get
$$-4a_{11} + 2a_{21} + 8a_{31} + 2a_{41} = 0$$
CONTRADICTION!! Since
$$-4a_{11} + 2a_{21} + 8a_{31} + 2a_{41} = 0 = -2 (2a_{11} - a_{21} - 4a_{31} - a_{41})= - 2T = -2 = 0 !!$$
Therefore NO SOLUTION
