Graph Theory adjacency matrix Is it possible for a graph to exist that meets these conditions?
For a Graph G, the adjacency matrix has all $1$’s in the first row and all $0$’s in the second row.
What I think: Such a graph cannot exist because if there were $1$’s in the first row, it means that v1 is connected to every single vertex in graph G. However, the second row being all $0$’s indicates that there is no vertex connected to v2, which is a contradiction?
Am I correct?
Also another question, with an adjacency matrix does a loop = 2 in the matrix?
 A: If you are dealing with an undirected graph, the adjcency matrix is a symmetric matrix. Hence, $a_{12} = a_{21}$. Hence, the adjacency matrix cannot represent an undirected graph.
However, if you are talking about adjacency matrix for a directed graph, then the second row can be zero, if there are no edges emanating from the second vertex (depending on the convention you choose).
A: 
Also another question, with an adjacacny matrix does a loop = 2 in the matrix?

This is a matter of definition, and as such, we get to choose.  The obvious possibilities are $1$ or $2$.  So which is best?
In this case, there is a good reason for writing $1$ for a loop in the adjacency matrix.  Namely, provided we write $1$ for loops in the adjacency matrix,  if $A=(A[i,j])$ is the adjacency matrix of a graph, then $A^k[i,j]$ is the number of walks from $i$ to $j$ of length $k$.
However, for the vertex's degree, it can be better to count a loop as contributing $2$, otherwise the Handshaking Lemma doesn't work (or, at least, would need adjusting).
