Determine the number of solutions as $a$ and $b$ change. Consider this first linear system.
\begin{cases}
ax_1+x_2=0\\
x_1+ax_2=0\\
2x_1+(1+a)x_2=a
\end{cases}
Using Gaussian Elimination on the associated matrix, I ended up with the following.
$$A =
\left[
\begin{array}{cc}
1&a&0\\
0&-2-a&a\\
0&1-a^2&0
\end{array}
\right]
$$
According to what I've done, the system should have a unique solution for $a= \pm1$ or $a = 2$, and none otherwise.
However the correct answer according to the book should be $a\neq\pm1$ for a unique solution, otherwise no solutions. I've studied the book cases and no matter how many times I try I can seem to solve these two systems correctly.
Consider now this second linear system.
\begin{cases}
4x_1+x_2=8\\
3ax_1-2x_2=0\\
5x_1+2x_2=5\\
-x_1+7bx_2=8
\end{cases}
Which I can't seem to solve correctly.
Here is one of my many attempts at solving the first system.

Please help and use $x$ and $y$ instead of $x_1$ and $x_2$ (so as not to make it awkward for you guys). Many thanks!
 A: The original system of equations:
$$
\begin{cases}
ax_1+x_2=0\\[4pt]
x_1+ax_2=0\\[4pt]
2x_1+(1+a)x_2=a\\
\end{cases} 
$$
The augmented matrix:
$$
\left[
\begin{array}{cc|c}
a&1&0\\
1&a&0\\
2&1+a&a\\
\end{array}
\right]
$$
Next, swap $r_1,r_2$ . . .
$$
\left[
\begin{array}{cc|c}
1&a&0\\
a&1&0\\
2&1+a&a\\
\end{array}
\right]
$$
Next, make the replacements:$\;r_2=r_2-ar_1\;\;\;$and$\;\;\;r_3=r_3-2r_1\;$. . .
$$
\left[
\begin{array}{cc|c}
1&a&0\\
0&1-a^2&0\\
0&1-a&a\\
\end{array}
\right]
$$
From the above, we see that if $a=1$, row $3$ would become $[0\;\;0\;\;1]$, hence for $a=1$, the system has no solutions.

So assume $a\ne 1$.

Next, swap $r_2,r_3$:
$$
\left[
\begin{array}{cc|c}
1&a&0\\
0&1-a&a\\
0&1-a^2&0\\
\end{array}
\right]
$$
Next, make the replacement:$\;r_3=r_3-(1+a)r_2\;$. . .
$$
\left[
\begin{array}{cc|c}
1&a&0\\
0&1-a&a\\
0&0&-a^2-a\\
\end{array}
\right]
$$
Next, for cosmetic improvement, make the replacement:$\;r_3=-r_3\;$. . .
$$
\left[
\begin{array}{cc|c}
1&a&0\\
0&1-a&a\\
0&0&a^2+a\\
\end{array}
\right]
$$
From the above, looking at row $3$, it follows that there are no solutions if $a^2+a\ne 0$.

It follows that for all values of $a$ exept $a=0, a=-1$, there are no solutions, and for those two values of $a$, the system has a unique solution (obtained using the first two rows of the reduced matrix).
