Given this system, find the values of $k$ for which this system has infinitely many solutions:
$$x+2y+z=2$$ $$2x-2y+3z = 1$$ $$x+2y-(k^2-3)z = k$$
After applying Gaussian elimination we arrive at the following augmented matrix:
$$[A \ \ b ] =\begin{bmatrix} 1 & 2 & 1 & 2 \\ 0 & -6 & 1 & -3 \\ 0 & 0 & -(k^2-3)-1 & k-2 \\ \end{bmatrix} $$
Now this system will have infinitely many solutions iff $$\left(-(k^2-3)-1 = 0\right) \land \left(k-2=0\right)$$.
But $\not \exists k \ni \left(-(k^2-3)-1 = 0\right) \land \left(k-2=0\right)$. Does that not imply that there doesn't exist a $k \in \mathbb{R}$, that would produce infinitely many solutions to this system? Have I made a mistake?, Are there other methods apart from Gaussian Elimination and trying to see if it produces a row full of zeros, to find out if a system will have no solution, a unique solution or infinitely many solutions?
Another way to is to check if there is any k such that rows of coefficients are not independent.It is very quick to find that when $k=\sqrt2$, row 1 and row 3 are not independent and thus can have infinite solution. I don't see $k=2$ has this features.