For this question:
$$ \begin{matrix} x & -2y & 3z = -1 \\ 2x & -4y & 6z = -2 \\ -x & 2y & kz = 1 \\ \end{matrix} $$
I ended up with this for the bottom two rows using the augmented matrix approach: $$ \left[ \begin{array}{ccc|c} 0&0&0&0\\ 0&0&(k+3)&0 \end{array} \right] $$
I got k = -3 but the answer is supposed to be that k has infinite solutions...
Whereas for this question I got k = -1 which is correct:
$$ \begin{matrix} x & -y & z = 3 \\ 2x & 3y & kz = 2 \\ 4x & 11y & -5z = 0 \\ \end{matrix} $$
How are these two questions different? Why is k infinite solutions in the first question and not the second? I used the augmented matrix approach to work this out so please use that method to explain to me how! Thanks.