Yes, Option (E) must be the right answer.
Number of edges in the graph:
Since the graphs are undirected, it can be observed that there will be two $1$'s in the adjacency matrix corresponding to each edge in the graph.
For example, suppose two there is an edge between nodes $A \ \& \ B$, then there will be $1$ in position $[A, B]$ & there will be a $1$ in position $[B,A]$ of the adjacency matrix.
That's why the given adjacency matrix is symmetric.
So the number of edges in the graph must be equal to half the number of $1$'s in the adjacency matrix.
Hence number of edges will be $7$ in the graph.
All the other graphs except (iii), have $7$ edges.So it is clear that the adjacency matrix does not represents graph (iii).
Isomorphism:
From the definition of Isomorphic graphs, it can be inferred that,
Isomorphic graphs must have same (adjacency matrix) representation.
Thus after eliminating graph (iii) we have to check for isomorphism among graphs (i), (ii) & (iv).
It can clearly be observed that graphs (ii) & (iv) are not isomorphic to each other.
It can also be observed that graph (i) & (ii) are isomorphic(Rotate graph (i) by $90$ degree left/right.
Graph (ii) is looking like a closed envelope in the figure, try to view it like an open envelope, like a trapezium over a rectangle.)
So now it can be inferred that either the adjacency matrix is representing both graphs (i) & (ii) or it is only representing (iv).
Cycles of length 6 :
Now from the adjacency matrix it can be observed that there should be a cycle of length $6$ in the graph, since $[1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 1] $ are all $1$'s in the matrix.(as $1$ at any position $[x, y]$ represents an edge between $x \ \& \ y$ in the graph).
& both graphs (i) & (ii) have cycles of length $6$, but graph (iv) does not has any cycle of length $6$, it has cycles of length $4 \ \& \ 5$ only.
Thus graph (iv) can not have the above adjacency matrix.
Hence the adjacency matrix represents graphs (i) & (ii).