I have to determine $c_1, c_2, c_3 \in \mathbb R$ with $c_1 < c_2$ that the coefficient matrix $(A_{c_1}|b_{c_1})$ and $(A_{C_1}|b_{c_2})$ do not have a result and for $(A_{C_{3}}|b_{C_{3}})$ infinite solutions.
$A_c=$ $$ \left( \begin{array}{cccc} 0&-2&4-5c-c^2\\ -2+c&1+c&2+3c+3c^2\\ 2-c&-c&-3+c-2c^2 \end{array} \right) $$
$b_c=$ $$ \left( \begin{array}{c} 1-c\\ c\\ 0\\ \end{array} \right) $$
So I have simplified the $Sol(A,b)$ matrix
$$ \left( \begin{array}{ccc|c} 0&-2&4-5c-c^2&1-c\\ -2+c&1+c&2+3c+3c^2&c\\ 2-c&-c&-3+c-2c^2&0 \end{array} \right) $$
to
$$ \left( \begin{array}{ccc|c} 2-c&-c&-3+c-2c^2&0\\ 0&1&-1+4c+c^2&c\\ 0&0&2+3c+c^2&1+c \end{array} \right) $$
to get the $c_3$ for infinite solutions:
$$2+3c+c^2=1+c$$ $$1+3c+c^2=c$$ $$1+2c+c^2=0, \Rightarrow c_3 = -1$$
Question: How can I get the values for $c_1$ and $c_2$?
My first guess was to write it down like that:
$x_1+x_3(-1+4c+c^2) = c$ $x_3(2+3c+c^2)=1+c$
and to simplify it somehow so there will be no solution for that. But how?