Solving matrices with unknown coefficients Thanks for reading. I've gone through the other thread on this topic but the answer is quite different to the one I've got for the following question and I need some help in checking if my answer is correct - any help is greatly appreciated :)
The question asks to solve for "k" to make the system:
(1) Consistent with many solutions  (2) Inconsistent
$$ \left[
    \begin{array}{ccc|c}
      1&1&k&6\\
      1&k&1&3\\
      k&1&1&7
    \end{array}
\right] $$
I got the following row reduction:
$$ \left[
    \begin{array}{ccc|c}
      1&0&-1&1/(k-1)\\
      0&1&-1&-3/(k-1)\\
      0&0&k+2&6 + 2/(k-1)
    \end{array}
\right] $$
Unfortunately I did not get the answer right. The answer is (1) none,  (2) -2 and 1
My justification for this is as follows:
From the last row: if K = -2, then 0 = [something] therefore system is inconsistent when K = -2.
From any row: if K = 1, then the solution will be 1/0 which doesn't exist therefore system is inconsistent when K = 1
Cannot have infinite solutions because no value of K will produce a Row of Zeroes.
I would appreciate any input on my answer/justification, especially if I've missed some vital concept.
Thank-you kindly! :)

The matrix row reduces to the following (without doing any 'division' by terms involving $k$:
\begin{bmatrix} 1 & 1 & k & 6\\ 0  & k-1 &  1-k & -3 \\ 0 & 0 & (k-1)(k+2) & 4-6k \end{bmatrix}
When doing row reduction, it is best not to divide by a term involving $k$ (if you divide by $k-1$ as you have above for example, then you need to consider the case $k=1$ separately). 
