Isomorphism of two graphs using adjacency matrix How can I show that the following two graphs are isomorphic:

Steps:
The given graphs can be written as:

 A: The adjacency matrix $A(G_2)$ is not symmetrical and one of the diagonal entry is non-zero. Most likely you have computed it wrongly.
To show isomorphism, it suffices to find permutation matrix $P$ such that $$PA(G_1)P^T=A(G_2).$$
The following observation might help you in constructing the matrix $P$.
$v_1$ corresponds to $w_5$. 
$v_2$ corresponds to $w_6$.
$v_3$ corresponds to $w_2$.
$v_4$ corresponds to $w_3$.
$v_5$ corresponds to $w_4$.
$v_6$ corresponds to $w_1$.
A: In this case it is easier to find the isomorphism by looking at the graphs rather than their adjacency matrices. Observe that in the first graph, there are three paths from $v_5$ to $v_3$, two of which have length 2 and one of which has length 3.  Similarly in the second graph from vertex $w_4$ to vertex $w_2$.  The paths of length 3 must be mapped to each other.  So, a bijection $f$ from the vertex set $\{v_1,\ldots,v_6\}$ of the first graph to the vertex set $\{w_1,\ldots,w_6\}$ of the second graph which preserves adjacency can be obtained.  
Once you have the bijection $f$, observe that it corresponds to a permutation $\pi \in S_6$, where $\pi$ maps $i$ to $j$ if $f$ maps $v_i$ to $w_j$.  Now, $\pi$ can be represented by a permutation matrix, and the adjacency matrix $A$ of the first graph and $B$ of the second graph satisfy the equation $PAP^T = B$.  
A: I am keeping this answer as simple as I can, so kindly pardon the layman's language. 
Observe the to graphs,
From G1, G2 can be obtained if the first line and second line of the graph G1 are interchanged. So, this graph is definitely iso-"morphic". (bijective and satisfies the edge adjacency property). And the mapping would be v6 -> w1, v1 -> w5 and v2 -> w6, rest being pretty straight forward mapping. These mappings satisfy the isomorphic property, hence it should be isomorphic.
