Why must $b=0$ for this linear system to have infinitely many solutions for all $a$? Consider the parameterized linear system of equations represented by the augmented matrix:
$$
\left[
\begin{array}{ccc|c}
1 & 0 & a & 1 \\ 
0 & 1 & 2 & 2 \\ 
0 & 0 & d-ab-2c & 3-b-2c
\end{array}
\right]
$$
I understand that to have infinitely many solutions, I need fewer rows than there are variables, and no inconsistency.  Therefore I see that $d-ab-2c=0$ and $3-b-2c=0$ are necessary to have infinitely many solutions.
But the question is more specific - it asks to find b,c,d that for all values of a, there are infinite solutions.  It's given in the answer that for this to happen, $b=0$ has to be added to the above two conditions.  The explanation I have looks like a tautology "if b is zero, it doesn't matter what a is, for every a we have infinite solutions", and I'd appreciate some insight into why this is so.
 A: As surmised in the Question, the augmented matrix represents a system with infinitely many solutions if and only if both $d−ab−2c=0$ and $3−b−2c=0$.  If the first of these expressions were nonzero, then the system would have a unique solution, and if it were zero while the second were nonzero, then the system would be inconsistent (have no solution).
So the original linear system can be set aside for now, and focus shifted to when $b,c,d$ would produce infinitely many solutions for all values of $a$.  That is, what fixed values of $b,c,d$ would produce:
$$ d−ab−2c=0 \;\text{ and }\; 3−b−2c=0 $$
regardless of what value $a$ is assigned.  Intuitively it is clear that for $a$ not to matter in this calculation, its coefficient $b$ must be zero, but we will show this with a proof.
In particular we must have the above conditions for both $a=0$ and $a=1$.  The first requires that we have $d-2c=0$ and $3-b-2c=0$.  The second requires $d-b-2c=0$ and $3-b-2c=0$.  Combining $d-2c=0$ and $d-b-2c=0$ implies $b=0$.
Once we know $b=0$, specific values for $c,d$ also follow.  The Reader should be able to deduce what these specific values are, from the conditions shown above.
