Elementary row operation, how does the determinant show up here?

Question

I'm stuck on a rather simple exercise..

Given the equation

we should find k as such that the system has either one, infinite or no solution.

Written as a matrix this translates to:

I understand the first row operation, which just switches the rows, but how to arrive at the third? And why is the determinant showing up here?

The teacher so far did not introduce determinants.

Answer 1

To arrive at the third, we did $R_{2}\rightarrow k\cdot R_{1}+R_{2}$ where $R_{i}$ is row $i$ .

So the system is now equivalent to $$x-ky=0$$ $$(1-k^2)y=0$$

So two cases to consider:

1. If $k\neq\pm1$ then $y=0$, so second is satisfied and the first gives $x=0$ (trivial solution)
2. If $k=\pm1$, then the second is satisfied for any $y$ and from the first we obtain $x=ky$ for $y\in\mathbb R$ (and we have infinitely many solutions). Clearly if $y=0$ then we are back at the trivial solution.

Answer 2

You take the first row times $k$ and add to the second row, so $1\cdot k + -k = 0$ and $-k \cdot k + 1 = 1-k^2$. That how you transform the matrix to row echelon form.