Explaining why a grammar is not LL(1) I need some help with explaining why a grammar is not LL(1).

Let us take the following grammar:

$$
\begin{align}
S \rightarrow &amp; aB \mid bA \mid \varepsilon \\
A \rightarrow &amp; aS \mid bAA \\
B \rightarrow &amp; b \\
\end{align}
$$

This is my attempt:

For the grammar to be LL(1) it is a necessary condition that for any strings $c_1γ$ and $c_2β$, derivable from $S \rightarrow aB$ and $A \rightarrow aS$ respectively, we have $c_1 \ne c_2$. 

But, $S \rightarrow aB$ and $A \rightarrow aS$, hence $c_1 = c_2$ and the grammar is not LL(1).

Is my reasoning right?

Thanks in advance.
If the grammar is ambiguous (at least one sentence has more than one parse tree), then the grammar is not in LL(1).

In general, you compute the LL(1) parse table and use it to answer the question. Your grammar is LL(1) if each cell in the table has at most 1 grammar rule. In other words, at least one table entry has more than 1 rule implies that your grammar is not in LL(1): http://en.wikipedia.org/wiki/LL_parser#Constructing_an_LL.281.29_parsing_table