Work out value of variables when given the sum of all combinations If I have a grid of the total of all possible combinations of the variables when they're multiplied, how can I work out the value of each individual variable?
For example:
Multiplication grid: https://i.imgur.com/BjPDXiJ.png
Those are the sums when each of the combinations of the variables are multiplied... how can I work out the value of each variable?
i.e:   $a = 5,\; b = 2,\; c = 7,\; d = 3,\; e = 2,\; f = 5,\; g = 3,\; h = 7$? 
 A: You must solve a system having $16$ equations and $8$ unknowns, so there is little hope to get a solution in general.
$
\left\{
\begin{array}{llll}
 a e=10 & b e=4 & c e=14 & d e=6 \\
 a f=25 & b f=10 & c f=35 & d f=15 \\
 a g=15 & b g=6 & c g=21 & d g=9 \\
 a h=35 & b h=14 & c h=49 & d h=21 \\
\end{array}
\right.
$
$$\left\{a,\frac{2 a}{5},\frac{7 a}{5},\frac{3 a}{5},\frac{10}{a},\frac{25}{a},\frac{15}{a},\frac{35}{a}\right\}$$
In this case you are lucky. There are infinite solutions. If you want all integer entries you must have $a=5$  and the other follow like this
$\{a=5,b=2,c=7,d=3,e=2,f=5,g=3,h=7\}$
A: Your system is standardly written as
$$
{\bf x}\;\overline {\bf y}  = {\bf x} \otimes \;{\bf y} = {\bf A}
$$
where the overline denotes the transpose, and the whole is known as a Dyad,
made from vectors $\bf x, \; \bf y$.    
Given $\bf A$, you want to solve for the vectors.
We shall premise some properties of the Dyad.
Calling $h$ the dimension of the vectors, if we multiply at the right by any of the $h-1$ vectors normal to $\bf y$
$$
{\bf 0} = \overline {\bf y} \,{\bf n}_{\,{\bf y}} \quad  \Rightarrow \quad {\bf 0} = {\bf x}\;\overline {\bf y} \,{\bf n}_{\,{\bf y}}  = {\bf A}\,{\bf n}_{\,{\bf y}} 
$$
we get a nullvector.
Same if we multiply $\overline {\bf A}$ to the left by the transposed of any vector normal to $\bf x$.
$$
{\bf 0} = \overline {{\bf n}_{\,{\bf x}} } \;{\bf x}\;\overline {\bf y} \, = \overline {{\bf n}_{\,{\bf x}} } \;{\bf A}
$$
Therefore $\bf A$ shall have a left and right null-space of dimension $h-1$, and 
consequently $\bf A$ shall have rank $=1$.
It is easy to see that the LU decomposition of $\bf A$ is given by
$$
{\bf A} = \left( {\matrix{
   {x_{\,1} } & 0 &  \cdots  & 0  \cr 
   {x_{\,2} } & 1 &  \cdots  & 0  \cr 
    \vdots  & {(0) \vdots } & {(1) \ddots } & {(0) \vdots }  \cr 
   {x_{\,h} } & 0 & {(0) \cdots } & 1  \cr 
 } } \right)\left( {\matrix{
   {y_{\,1} } & {y_{\,2} } &  \cdots  & {y_{\,h} }  \cr 
   0 & 0 &  \cdots  & 0  \cr 
    \vdots  &  \vdots  &  \ddots  &  \vdots   \cr 
   0 & 0 &  \cdots  & 0  \cr 
 } } \right)
$$
and that the only generally non-null eigenvalue is equal to ${\bf x} \cdot {\bf y}$, which
is then null as well if the two vectors are normal.
In conclusion, provided that the rank of the matrix be $1$, you can recover the vectors by its LU decomposition (or from the nullspace, or eigenspace).
Of course, as underlined by achille hui, the vectors you get are well determined, less a constant multiplier/divisor.
