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.
Matrix of coefficients is $$A=\left( \begin{array}{ccc} 1 & k & -1 \\ 2 & -1 & k \\ 1 & 10 & -6 \\ \end{array} \right)$$ $\det A=k^2+2 k-15$
If $\det A\ne 0\to k\ne 3;\;k\ne-5$ the system has one and only one solution.
If $k=3$ then $\det A=0$ and $\text{rank}(A)=\text{rank}(A|B)=2$ thus there are infinite solutions
$$A|B=\left( \begin{array}{ccc|c} 1 & 3 & -1 & 2 \\ 2 & -1 & 3 & 5 \\ 1 & 10 & -6 & 1 \\ \end{array} \right) $$ If $k=-5$ $$A|B=\left( \begin{array}{cccc} 1 & -5 & -1 & 2 \\ 2 & -1 & -5 & 5 \\ 1 & 10 & -6 & 1 \\ \end{array} \right)$$ $\text{rank}(A|B)=3$ while $\text{rank}(A)=2$. As ranks are different, the system has no solutions.
see here for Rouchè-Capelli theorem
