I've been given the following adjacency matrix:
$$\left(\begin{array}{cccccccc} 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \\ 1 & 1 & 0 & 0 & 0 & 1 & 1 & 0 \\ \end{array}\right)$$
and I need to prove that it is a self complementary graph. All this must be done by hand so I found the complementary graph but I don't know how to continue.
Do I need to solve
$$Ax=b$$
or
$$x^{t}Ax=b$$
is $x$ a matrix defined by 8 columns and 8 rows?
The blue graph has the big square 2-4-6-8 with little triangles 2-3-4, 4-5,6, 6-7-8, and 8-1-2, and also the little triangle vertices which are opposite relative to the big square are joined, i.e. 1-5 and 3-7.
The red graph has a similar description: there is a big square 1-3-5-7, and triangles 1-6-3, 3-8-5, 5-2-7, and 7-4-2, and also the triangle vertices which are opposite relative to the big square are joined, i.e. 2-6 and 8-4.
This means one can set up an isomorphism by corresponding the vertices as
Blue: $[1,2,3,4,5,6,7,8],$
Red: $[4,1,6,3,8,5,2,7].$