The system
$$\begin{cases}x-y+3z=-5\\5x+2y-6z=\alpha \\2x-y+\alpha z = -6 \end{cases}$$
for which $\alpha$ values the linear equation system:
- has no solution
- has one solution
- has more than one solution
I started to do Gauss elimination on it, but i have no idea what i am looking for and how to approach this, I'm stuck with the Gauss elimination.
My work so far: \begin{align} \left(\begin{array}{rrr|r} 1 & -1 & 3 & -5 \\ 5 & 2 & 6 & \alpha \\ 2 & -1 & \alpha & -6 \end{array}\right) &\leadsto \left(\begin{array}{rrr|r} 1 & -1 & 3 & -5 \\ 0 & 7 & -9 & \alpha + 25 \\ 2 & -1 & \alpha & -6 \end{array}\right) \\ &\leadsto \left(\begin{array}{rrr|r} 1 & -1 & 3 & -5 \\ 0 & 7 & -9 & \alpha+25 \\ 0 & 1 & \alpha-6 & 4 \end{array}\right) \\ &\leadsto \left(\begin{array}{rrr|r} 1 & 0 & \alpha + 3 & -1 \\ 0 & 7 & -9 & \alpha+25 \\ 0 & 1 & \alpha-6 & 4 \end{array}\right) \\ \end{align}