Linear Algebra Question: get amount each person tipped from same amount of central tip jars Essentially, there are six guests and six tip jars (1 between each pair of guests). If we know the amount in each tip jar, and know each guest gives half their money as a tip, can we figure out the starting balances of each guest? Here is the full question text:
Suppose six guests (represented by green circles) sit around a hexagonal table and there are six jars of
tips (represented by black dots). If we know the amount of tip in each jar, J1 to J6, can we determine
each individual’s tip amount, G1 to G6? If yes, explain why by examining the relationship between the
jar values, J1 to J6, and guest tips, G1 to G6. If not, give two different assignments of G1 to G6 that
will result in the same J1 to J6.

I think you want to set this up as:
\begin{pmatrix}0.5&0.5&0&0&0&0&a\\ 0.5&0&0.5&0&0&0&b\\ 0&0.5&0&0.5&0&0&c\\ 0&0&0.5&0&0.5&0&d\\ 0&0&0&0.5&0&0.5&e\\ 0&0&0&0&0.5&0.5&f\end{pmatrix}
Where a,b, etc. are the guest balances. But I'm unsure where to progress from here.
I'm also interested how this changes if the number of guests is 5.
Any help is appreciated
Sincerely,
D
 A: Since the unknowns are the guest balances you should not write them in the last column (the "solution column" of the linear system of equations) since the unknowns of a system of the kind $Ax=b$ are the components of $x$. So in this cas $b= (J1, ..., J6)$ and $x=(G1,..., G6)$ and the matrix you write down to solve the system of equations is $(A |b)$. So the values of the last column should be the jar balances J1 to J6 and each column represents one of the guests and how this guest splits up his/her tip.
The resulting equations are (if J1 ist between Guest 1 and Guest, J2 is between Guest 2 and Guest 3 and so on...):
$$J1 = 0,5 \cdot G1 + 0,5 \cdot G2$$
$$J2 = 0,5 \cdot G2 + 0,5 \cdot G3$$
Putting it all together in a matrix where G1 to G6 are the unknowns we get:
$$\begin{pmatrix} 0,5&0,5&0&0&0&0&J1\\
0&0,5&0,5&0&0&0&J2\\
0&0&0,5&0,5&0&0&J3\\
0&0&0&0,5&0,5&0&J4\\
0&0&0&0&0,5&0,5&J5\\
0,5&0&0&0&0&0,5&J6
\end{pmatrix}$$
The matrix looks like this becaus Guest 1 puts his tip in the Jars 1 and 6
For any other number $n$ of guests we will get a similar looking matrix $A \in M_{n \times (n+1)}$ where the last column is the "solution column" with the jar balances.
Hope this explanation helps you.
