Given the linear system:
$$\begin{cases} x_1 + kx_2 − x_3 = 2\\ 2x_1 − x_2 + kx_3 = 5 \\ x_1 +10x_2 −6x_3 =1 \end{cases}$$
for which values of $k$ has the system (2): (a) No solutions (b) A unique solution. (c) Infinitely many solutions.
I've been trying echelon form where i switched $R_1$ with $R_3$ and then i switched $R_2$ with $R_3$
So I have $\left[\begin{array}{ccc|c}1&10&-6&1\\1&k&-1&2\\2&-1&k&5\end{array}\right]$
but then I'm stuck and don't know how to get any further.