Problem: Solve the following system in function of the parameter $b$:
\begin{align*} \begin{cases} -bx + 2y - (2+b^2)z + bu &= -2 \\ x -2y + bz -u &= 0 \\ x + (2b-4)y + (2-b)z + (b-1)u &= 2 \\ x -2by -(3b+2)z + (4b-5)u &= 2b-4 \end{cases} \end{align*}
Attempt at solution: We write down the augmented matrix of this system, and then apply the operations: $R_1 \leftrightarrow R_2, R_2 \rightarrow R_2 + b R_1$. This gives us the matrix: \begin{align*} \left(\begin{array}{cccc|c} 1 & -2 & b & -1 & 0 \\ 0 & (2-2b) & (2+b^2) & 0 & -2b \\ 1 & (2b-4) & (2-b) & (b-1) & 2 \\ 1 & -2b & -(3b+2) & (4b+5) & (2b-4) \end{array}\right) \end{align*} After that, we do $R_3 \rightarrow R_3 - R_1$ and $R_4 \rightarrow R_4 - R_1$: \begin{align*}\left(\begin{array}{cccc|c} 1 & -2 & b & -1 & 0 \\ 0 & (2-2b) & (2+2b^2) & 0 & -2b \\ 0 & (2b-2) & (2-2b) & b & 2 \\ 0 & (-2b+2) & (-4b-2) & (4b-4) & (2b-4) \end{array}\right) \end{align*} Now I want a leading $1$ at the position $a_{22}$.
Case 1. Let $b = 1$. Then our matrix reduces to \begin{align*} \left(\begin{array}{cccc|c} 1 & -2 & 1 & -1 & 0 \\ 0 & 0 & 4 & 0 & -2 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & -6 & 1 & -2 \end{array}\right) \end{align*} From the second row we see than that $u =2$. Substituting this in the last equation gives $z = - \frac{2}{3}$. But from the second row $z = - \frac{1}{2}$, which is a contradiction. Does this mean I can conclude the system has no solutions in this case?
Then if $b \neq 1$, should I just proceed with Gauss-elimination untill I hit another case distinction?