Ax=b with parametric b vector. A=$\begin{pmatrix}
1 & 3 & 1\\ 
0 & 1 & 1\\ 
2 & -2 & -6
\end{pmatrix}$
B=$\begin{pmatrix}
8\\ 
k\\ 
8
\end{pmatrix}$
Discuss the solutions of the system S: Ax=B.

I have used the Rouche Capelli theorem wich implicate that S have solutions only if:
$rkA\leq$rk$A'$     
where A'= A|B.
Well, $rkA$=2 still. But $rkA'=2$ only if $\begin{vmatrix}
1 & 3 & 8\\ 
0 & 1 & k\\ 
2 & -2 & 8 
\end{vmatrix}$=$0$
This happens for $k=1$ wich is ok with my textbook but...   ...if $k=1$ I can also calculate the other possible determinant:
$\begin{vmatrix}
1 & 1 & 8\\ 
0 & 1 & k\\ 
2 & -6 & 8 
\end{vmatrix}$ wich for $k=1$ is not equal to $0$ $\Rightarrow $ for $k=1$,  $rkA'$=$0$
My textbook still saying that for $k=1$, S have one solution. Why i'm wrong?
My solution will be to create a system made of the possible determinant equalized to 0 (in this case 2 deterimnants).

Use row reduction for the augmented matrix:
\begin{align}
\begin{bmatrix}1&3&1&8\\0&1&1&k\\2&-2&-6&8\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&3&1&8\\0&1&1&k\\0&-8&-8&-8\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1&3&1&8\\0&1&1&k\\0&0&0&(k-1)8\end{bmatrix}\end{align}
Thus $\operatorname{rank}A'=2$ if and only if $k=1$.
Moreover, the solutions are an affine subspace of dimension $1$: if we continue row reduction to obtain the reduced row echelon form, dropping the last (zero) row,we have
$$\begin{bmatrix}1&3&1&8\\0&1&1&1\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&-2&5\\0&1&1&1\end{bmatrix}$$
whence the solutions:
\begin{align}
\begin{cases}x&=2z-5,\\
y&=-z-1,\end{cases}\quad\text{or, in vector form:}\quad \begin{bmatrix}x\\y\\z\end{bmatrix}=z\begin{bmatrix}2\\-1\\1\end{bmatrix}-\begin{bmatrix}5\\1\\0\end{bmatrix}.
\end{align}
