Why is this the $L(M)$ of this DFA? Why is this the $L(M)$ of this DFA? Can someone please explain it? I am new to this course. When I tried alone answering the question of "What is special about the words that get accepted by this DFA?" I answered that "any word that ends with a while the letter before it is not a ($\epsilon a$ for example or $ba$ or...."


First of all note that your DFA accepts the string $\newcommand{\aa}{\mathtt{a}}\newcommand{\bb}{\mathtt{b}}\aa\aa\aa$, so your characterisation of the language it accepts is a bit off.
Hint.  An inspection of the DFA reveals the following simple facts:


*

*Reading an $\aa$ from either $q_1$ or $q_3$ brings you to the accepting state.

*Reading a $\bb$ from either $q_1$ or $q_3$ brings you to the opposite state.

*Reading either an $\aa$ or a $\bb$ from the accepting state $q_2$ brings you to one of $q_1$ or $q_3$.


Concentrate first on the ways you can get from the accepting state back to itself (in particular by only reading consecutive $\aa$s), and then note that the only way to get from a non-accepting state to the accepting state is to read an $\aa$.
