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!
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).