# 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?

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You can just assume that det A is nonzero, and then can solve for x,y,z using Cramer's rule, for example. –  coffeemath Nov 27 '12 at 22:22
but if a=1, the det A = 0, –  amWhy Nov 27 '12 at 22:23
Yes coffemath, but I need to discuss also the case $-2+b+ab = 0$. –  user50554 Nov 27 '12 at 22:26
See my answer below in which I discuss the case $-2+b+ab=0$. –  coffeemath Dec 2 '12 at 11:36

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 $$x = \frac{ (a-b)(b-1)}{(a-1)(ab+b-2)},$$ $$y=\frac{(ab+a-2)(b-1)}{(a-1)(ab+b-2)b},$$ $$z=\frac{a-b}{ab+b-2}.$$

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