I may be wrong but from the matrix example you found, I can say that using matrices is not for finding the isomorphism directly but checking (we can say proving as well) whether the isomorphism we found is correct or not. Because in your example, notice that s/he already knows the isomorphism between graphs as "(A B C D E F G) = (7 4 3 6 5 2 1)".
In the case of Petersen graph (the graph on the left), actually you can find an isomorphism if you can be more systematic. First of all, let us say $G$ is the Petersen graph (on the left) and $H$ is the graph on the right.
There is a graph isomorphism $\theta: G \rightarrow H$ if and only if
for any adjacent vertices $x,y \in V(G)$, $\theta(x)$ and $\theta(y)$
is adjacent in $H$ and for any non-adjacent vertices $u,t \in V(G)$, $\theta(u)$ and $\theta(t)$ are non-adjacent in $H$.
I am writing the definitions because your answer is not correct and you can observe it by using the definition. For example, you map $b \rightarrow 1$ and $g \rightarrow 4$. In $H$, notice that vertices $1$ and $4$ are adjacent but in $G$, vertices $b$ and $g$ are not adjacent. So your mapping does not give an isomorphism. Instead of trying it randomly, you can use the symmetry of the graphs.
Let's start with the vertex $a$ and let's say $\theta(a) = 2$, that is $a \rightarrow 2$ (After this example, I will always use $\theta$ representation). Then notice that vertices $b$, $d$ and $f$ are adjacent to $a$. Let's say $\theta(b) = 1$, $\theta(f) = 3$ and $\theta(d) = 5$. From here, I suggest you to choose your next vertex from one of the vertices that you have mapped already. So let's choose the vertex $d$ and find its adjacent vertices. Those are $a$, $g$ and $h$. We have already mapped $a$, so for $g$ and $h$, we have only two options (Because in $H$, $5$ is adjacent to $2$, $7$ and $8$ and we have mapped $2$ already). So let's say $\theta(g) = 7$ and $\theta(h) = 8$.
Now for the rest, remember that we have $b \rightarrow 1$ and $h \rightarrow 8$. Notice that there is only one vertex adjacent to both $b$ and $h$ in $G$ (or $1$ and $8$ in $H$) and that is $c$ in $G$ (or $10$ in $H$). So we can map $\theta(c) = 10$. After this part, by a similar method, you can find other mappings as the following (with previous ones):
$$\theta(a) = 2,\ \theta(b) = 1,\ \theta(c) = 10,\ \theta(d) = 5,\ \theta(e) = 9,\ \theta(f) = 3,\ \theta(g) = 7,\ \theta(h) = 8,\ \theta(i) = 4,\ \theta(k) = 6$$
Now we can prove that $\theta$ is an isomorphism by using the matrices. I just constructed two tables as matrices and I will put them on because how the matrices are constructed has already been explained in the example you have found. In this case, first matrix will have indexing as $a,b,c,d,e,f,g,h,i,k$ and the second matrix will have indexing as we found in isomorphism, that is, $2,1,10,5,9,3,7,8,4,6$. According to these indexing, the matrices will be as the following and they are the same if you reconstruct them from the start:
Since they are the same, we can conclude that $\theta: G \rightarrow H$ is a graph isomorphism.
