$$u = (r,1,1), \ v = (1,s,2s), \ w = (1,2,2)$$
I know to assume the set will be linearly dependent, equate them to the $0$ vector and find the values not equal to what $r$ and $s$ is. But I can't seem to reduce the matrix to echelon form.
I've gotten to the point of:
$$ \begin{pmatrix} r & 1 & 1 \\ 1 & s & 2 \\ 1 & 2s & 2 \\ \end{pmatrix} $$ $R_2 = R_2 - R_3$ $$ \begin{pmatrix} r & 1 & 1 \\ 1 & -s & 0 \\ 1 & 2s & 2 \\ \end{pmatrix} $$ $R_3 = R_3 +2R_2$
$$ \begin{pmatrix} r & 1& 1 \\ 0 & -s & 0 \\ 1 & 0 & 2 \\ \end{pmatrix} $$
EDIT: Thank you @LordSharktheUnknown
$R_3 = r \times R_3 - R_1$
$$ \begin{pmatrix} r & 1 & 1 \\ 0 & -s & 0 \\ 0 & -1 & 2r-1 \\ \end{pmatrix} $$
And I'm stuck. I know $s ≠ 0$ for it to be linearly independent but I have no idea how to reduce the rows any further to get to $r$ by itself.
Any help is appreciated. Thank you.
Take the last pre-edit matrix (the edit adds a transformation that changes linear dependence in case $r=0$; it's also not the transformation Lord Shark the Unknown describes).
By development in the second row (is that the correct English term?), one easily calculates the determinant as $$\det \begin{pmatrix} r & 1& 1 \\ 0 & -s & 0 \\ 1 & 0 & 2 \\ \end{pmatrix} = -s(2r-1)$$ Now the vectors are linearly independent iff the determinant is non-zero, which is the case iff all factors are nonzero. Thus we easily see the conditions: $$s\ne 0, r\ne \tfrac12$$