Lambda productions in grammar I tried removing the $\lambda$ productions from this grammar:
$S \rightarrow a A b \mid B B a$
$A \rightarrow b b \mid \lambda$
$B \rightarrow A A \mid b A a $
It seems like you just take away the 2nd part of the $A$ production since that's the only relevant here and then the rest of it is terminal one way or the other. But it seems like there's more to it:
$S \rightarrow a A b \mid B B a$
$A \rightarrow  b b$
$B \rightarrow A A \mid b A a$
I think I'm on the right track here but not really 100% sure.  Some guidance would be helpful.
 A: While it’s true that you will eventually eliminate the production $A\to\lambda$, you can’t just throw it away: by doing so, you’ve changed the language generated by the grammar. The original grammar allows the derivation $$S\Rightarrow aAb\Rightarrow ab\;,$$ while your new one cannot generate the word $ab$ at all. To see this, note that every one of your productions increases the length of the derived string, and the only terminal production is $A\to bb$; it takes at least one other step just to get an $A$, so your grammar cannot possibly produce any word of length less than $3$. (In fact the shortest possible length is $4$, for $abbb$.)
The idea is to adjust all productions that have $A$ on the righthand side in a way that compensates for losing the production $A\to\lambda$. If we have a production $X\to vAw$, where $v$ and $w$ are any strings of terminal and/or non-terminal symbols, the production $A\to\lambda$ allowed the derivation $X\Rightarrow vAw\Rightarrow vw$. If we throw out the production $A\to\lambda$, we lose this possibility, so to compensate we add the production $X\to vw$; this permits the derivation $X\Rightarrow vw$, which has the same effect as the original derivation $X\Rightarrow vAw\Rightarrow vw$ that is no longer available.
In particular, this means that we must add the production $S\to ab$ alongside the existing $S\to aAb$ and the production $B\to ba$ alongside the existing $B\to bAa$.
Taking care of $B\to AA$ is a little trickier, but it’s still not bad if you think in terms of compensating for loss of $A\to\lambda$. In the original grammar we have derivations 
$$\begin{align*}
&B\Rightarrow AA\Rightarrow A\lambda=A\text{ and}\\
&B\Rightarrow AA\Rightarrow\lambda A=A
\end{align*}$$
that are no longer available when we throw out $A\to\lambda$, so we have to add $B\to A$. But we also have
$$B\Rightarrow AA\Rightarrow^*A\Rightarrow\lambda\;,$$
in which we apply the $\lambda$ production to both $A$’s, so we also need to add $B\to\lambda$. At this point we have the following productions:
$$\begin{align*}
&S\to aAb\mid ab\mid BBa\\
&A\to bb\\
&B\to AA\mid A\mid\lambda\mid bAa\mid ba\;.
\end{align*}$$
We got rid of $A\to\lambda$, but at the cost of introducing a new $\lambda$ production, $B\to\lambda$. That’s okay: we can repeat the process to get rid of $B\to\lambda$. The only production with $B$ on the righthand side is $S\to BBa$. The derivations that we lose by throwing away $B\to\lambda$ are $$S\Rightarrow BBa\Rightarrow Ba$$ and $$S\Rightarrow BBa\Rightarrow Ba\Rightarrow a\;,$$ so we can compensate for the loss of $B\to\lambda$ by adding productions $S\to Ba$ and $S\to a$:
$$\begin{align*}
&S\to aAb\mid ab\mid BBa\mid Ba\mid a\\
&A\to bb\\
&B\to AA\mid A\mid bAa\mid ba\;.
\end{align*}$$
A: From this paper:

  
*
  
*Find nonterminals that derive $\lambda$.
  
*For each production $A \rightarrow w$ construct all productions $A \rightarrow w’$ where $w’$ is obtained from
  $w$ by removing one or more occurrences of the nonterminals from Step 1.
  
*Combine the original productions with those of step 2 and eliminate any $\lambda$-productions.
  

$S \rightarrow a A b | B B a$
$A \rightarrow bb|\lambda$
$B \rightarrow AA | bAa$


*

*$A$ derives $\lambda$.

*$S \rightarrow a b$ 
$A \rightarrow bb$
$B \rightarrow A | \lambda | ba$

*$S \rightarrow aAb | BBa | ab$ 
$A \rightarrow bb$
$B \rightarrow AA | A | \lambda | bAa |ba$
Note that since you still have a $\lambda$-production in $B$, you will need to apply this algorithm again to obtain your final answer.
