Given this linear system for $x,y,z \in \mathbb{R}$
\begin{aligned} (a+1)x + ay - az = 2+a \\ (a+1)x + (a+2)y - (a + 2)z = a+4 \\ (a+1)x+ ay +(a^2-6)z = a^2 - 2a + 4 \\ (2a+2)x+ 2ay +(a^2 - a - 6)z = a^2 - a + 6 \\ \end{aligned} For what values of $a$ does the system have one solution? infinite solutions? no solutions?
I proceed by putting its associated augmented matrix in row reduced echelon form and get that $x = \frac{2}{a+1}$, $y = 1 + \frac{(a-1)(a-2)}{(a+3)(a-2)}$ and $z = \frac{(a-1)(a-2)}{(a+3)(a-2)}$. I understand that if $a = -3 $ or $a = -1$ that the system will have no solutions. however I am confused about $a = 2$. If i plug in a = 2 into the original system and try to row reduce from there, I get infinite solutions. If i plug $a = 2$ into my solutions for $x,y,z$ and cancel out the $(a-2)$ terms, and plug these back into the original system, they seem to work out perfectly. So whats going on with $a = 2$? I haven't found anything in my book about how to deal with this. Am I allowed to cancel out the $\frac{a-2}{a-2}$ in my solutions for $z$ and $y$? Any help in the right direction would be appreciated.