Let there be the following system of linear equations: $$x+z+bw=a \\ ax+y+az+(a+ab)w=1+a^2 \\ bx+(a+b)z+(1+b^2)w=4+a\\ bx+bz+(a-ab+b^2)w=a+1+ab$$
a,b parameters. The question is, for which a,b there is no solution to the system, for which there are infinite and for which there is one. I put it into a matrix and with some row operations I got to: $$\begin{array}{cccc|c} 1 & 0 & 1 & b&a \\ 0&1&0&a&1\\ 0&0&a&1&4\\ 0&0&0&a-2b&a+1 \end{array}$$
How do I continue from here? I'm quite confused. For instance I thought that saying that for all b, if a=2b then there is no solution. But that doesn't sound quite right...
Thanks in advance for any help!
You've done good until that row reduced form (I assume the calculations are correct).
Now, if $a=0$, you can do another row operation: $$ \left[\begin{array}{cccc|c} 1 & 0 & 1 & b&0 \\ 0&1&0&0&1\\ 0&0&0&1&4\\ 0&0&0&-2b&1 \end{array}\right] \to \left[\begin{array}{cccc|c} 1 & 0 & 1 & b&0 \\ 0&1&0&0&1\\ 0&0&0&1&4\\ 0&0&0&0&1+8b \end{array}\right] $$ What can you say about this case?
If, instead, $a\ne0$, you have to distinguish when $a-2b=0$ or not.
If $a-2b=0$ you have $a=2b$ and you have a solution if and only if $a+1=0$, that is $a=-1$ and $b=-2$.
If $a-2b\ne0$, then …