<s>[INST] Best way to discuss a system of linear equations that depends on two parameters. I know there is no specific way, but what is, in your opinion, the best way to discuss a system of linear equations like:
$$ 
\left\{
\begin{array}{l}
ax+by+z = 1\\
x+aby+z=b\\
x+by+bz=1
\end{array}
\right.
$$
$$(A|B) = \left(\begin{array}{ccc|c}
a & b & 1 & 1\\
1 & ab & 1 & b\\
1 & b & b & 1
\end{array}
\right)
$$
If $A$ is the coefficient matrix, $\det A = b(a-1)(-2+b+a b)$, so it is a little bit hard to discuss the system using determinants, right?
 [/INST] If $b=0$ the system leads to also $a=2$ and then the solution is $x=1,z=1$ with $y$ arbitrary.
If $a=1$ it leads to also $b=1$ and in this case the system becomes the single relation $x+y+z=1$.
Now if we assume that $b$ is not zero (since the case $b=0$ already dealt with), but that $-2+b+ab=0$, then we have $a=(2-b)/b$ [OK since $b$ nonzero]. In this case the system is inconsistent unless it happens that $b^2+b-2=0$, i.e. $b=1,-2$. The case $b=1$ leads again to simply $x+y+z=1$, while the case $b=-2$ has also $a=-2$ and the solution may be written $x=-z$, $y=-1/2-1/2z$, and $z$ arbitrary.
In the remaining case where the determinant is nonzero, Cramer's rule can be used. It leads to
$$x = \frac{ (a-b)(b-1)}{(a-1)(ab+b-2)},$$
$$y=\frac{(ab+a-2)(b-1)}{(a-1)(ab+b-2)b},$$
$$z=\frac{a-b}{ab+b-2}.$$
</s>