$n \times m$ matrix conversion? Is it possible to convert an $n\times m$ matrix $A$ such that
$$ A=CB $$
where $B$ is a $1\times m$ matrix which contains all elements of $A$, and $C$ is a $n\times 1$ matrix. I'm assuming no since this might give a special case of matrices.. but i am not so sure. If this is not possible, is it possible to extend matrix $A$ to ($n$ by $m$) by ($n$ by $m$) so the same conditions are met, yet the matrix is replicated and the result needs to be unique. Just to give a reason for this, i figured out a way to make $A$ into a $1$ by ($n$ by $m$) vector $B$, but to find an inverse of this, i need to solve $A=CB$, which is what's giving me problems. 

assume ellements of $C$ and $B$ in variable form like $C_i$ and $B_j$ where $i$ represents row and $j$ represents column. now for any $A_{ij}=C_i\cdot B_j$ thus lets say for first row $A_{1j}=C_1\cdot B_j$ thus ratio of the elements of $B$ is the same as the ratio of first row elements of $A$ but wait $A_{2j}=C_1\cdot B_j$ 
thus its a contradiction , unless all row have same ratios of their elements it is impossible to form $A=CB$ as for the second part i can always take $C=kA$ and $B=\frac{I}{k}$ where $k$ is a constant and $I$ is the identity matrix.
