Weighted/Directional Graph and associated Adjacency Matrix This is a question from Intellipath, a program my school uses to teach lessons...
I apologize if this is a type of redundant question, but my question has to do with the graph specifically. I understand weighted and directional graphs just fine, I also understand adjacency matrixes just fine. However, according to my instructor and this program, I am wrong in this case, but I don't see it. I believe that Towns(4,1) = 12, and the opposite (1,4) is -12, 0, or the infinite symbol (don't know how to type it). Either way, (4,1) is equal to 12, that is the important one. My instructor is telling me that (4,1) is -12 and I do not understand where he is getting that. He has asked other collegues of his and they supposedly agree, but again, I don't see it. I have asked many friends of mine, googled other examples and have reached out to other sites, and I have yet to find someone who disagrees with me, but I figured I would try here as well. So, IF somehow, (4,1) = -12, please explain that to me....thank you for your time.
 A: It is purely a matter of convention: it is clear that, in words, the distance from $D$ to $A$ is $12$. We all agree on this. The question is: how do you represent this information as an entry of the adjacency matrix? Do you represent the distance from $D$ to $A$ as $T(4,1)$ or as $T(1,4)$? It's a matter of convention, but the more intuitive option would be to use $T(4,1)$ here, which means that you are right.
Concerning the minus sign, it's again a matter of convention: given that there is no arrow pointing from $A$ to $D$, I interpret this as saying that you cannot go directly from $A$ to $D$, which symbolically means that the distance from $A$ to $D$ is $\infty$, i.e. (using the convention chosen in the above paragraph), $T(1,4) = \infty$. The problem is that there is no way to place such an abstract symbol as $\infty$ in a computer program, therefore it is better to represent it as negative values: every negative entry should be read as meaning the absence of an oriented edge between the corresponding vertices. Using $0$ to encode this absence is not a good idea: sometimes $0$ is used on the diagonal of the matrix, representing the fact that one may go from any town $X$ into $X$ at no cost at all (but, again, this is yet another convention).
As you see, everything depends on conventions, and one cannot give a more detailed answer without knowing which ones your instructor uses. In any case, if your instructor uses "the other" convention regarding the order of the indices (i.e. $T(1,4)$ instead of $T(4,1)$), you shouldn't worry too much, because his matrix will be the transposed of yours, and all the significant theorems about oriented graphs do not "feel" this transposition (i.e. they remain true and give the same results with and without transposition).
A: I think the problem is how you are interpreting what each cell in your table represents.
The cell that represents going from D to A should be 12 and the cell that represents going from A to D should be -12, based on the digraph presented and assuming that traveling against the direction is interpreted as negative.
You and your instructor have to be clear on which cells represents which travels.
