What values of $a$ and $b$ does this system have infinitely many solutions? As a disclosure, this question is more for me to confirm that I did my work correctly. More specifically, the "solution" provided to me claims there are two values of $a$ and $b$ that yield infinite solutions, but I found only one. 
That said, this is the problem:
For what values of $a$ and $b$ does this system have infinitely many solutions 
\begin{equation} A =
\begin{pmatrix}
    a & 0 & b & |2 \\
    a & 2 & a & |b \\
    b & 2 & a & |a \\ 
\end{pmatrix}
\end{equation}
I reduced the matrix to 
\begin{equation} A =
\begin{pmatrix}
    a & 0 & b & |2 \\
    0 & 2 & a-b & |b-2 \\
    0 & 0 & b-\frac{b^{2}}{a} & |2-b+a-2\frac{b^{2}}{a} \\ 
\end{pmatrix}
\end{equation}
For this system to have infinite solutions, then one row must be $0$. In order for that to happen, I determined that $b=0$, $a=-2$.
However, the solution I have claims $b=0$, $a=-2$, or $a = b \neq 0$. I don't see how the latter is possible based on my reduction. Either I did something wrong, or the solution is wrong. Any insight would be appreciated.

You can check easily if the solution is wrong. (It is not.) Put $a=b$. Then you have: 
\begin{equation} A =
\begin{pmatrix}
    a & 0 & a & |2 \\
    a & 2 & a & |a \\
    a & 2 & a & |a \\ 
\end{pmatrix}
\end{equation}
Reducing to (III - I) and (II - I)
\begin{equation} A =
\begin{pmatrix}
    a & 0 & a & |2 \\
    0 & 2 & 0 & |a-2 \\
    0 & 0 & 0 & |0 \\ 
\end{pmatrix}
\end{equation}
For $a\neq 0$ this has infinitely many solutions. (For $a=0$ it has none.) 
Thus there is a problem in your computation.  
