Linear combinations of vectors that depend on a parameter 
Let $s \in \mathbb{R}$ and
  $$\mathbf{u} = \begin{bmatrix} 1 \\ 4 \\ 0 \end{bmatrix}, \quad
\mathbf{v}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad
\mathbf{w}=\begin{bmatrix} s^2 \\ 9 \\ s^2 \end{bmatrix}$$
  for which values $s \in \mathbb{R}$ is
  $$\mathbf{b} = \begin{bmatrix} 4+2s \\ 10 \\ 2s \end{bmatrix}$$ a linear combination of $\mathbf{u},\mathbf{v}$ and $\mathbf{w}$?

Through elementary row operations I've reduced the matrix to 
\begin{bmatrix}
1 & 0 & 0 & 4\\
0 & 1 & s^2 & 2s\\
0 & 0 & \frac{s^2-9}{9} & \frac{1}{9}(2s+6) \\
 \end{bmatrix}
I'm not too sure how to progress from here? I think this may be the next step:
$$\begin{bmatrix} x_1 \\ x_2+x_3 \\ x_3 \end{bmatrix} = \mathbf{b} = \begin{bmatrix} 4+2s \\ 10 \\ 2s \end{bmatrix}$$
then say that a solution is $s=0$, but it asks for values as in more than one, which I can't seem so see.
 A: Note that $$\mathrm{det}\left(\begin{array}{rrr} 1 & 1 & s^2 \\ 4 & 1 & 9 \\ 0 & 1 & s^2\end{array}\right)=s^2-9.$$ Thus, if $s\ne \pm 3,$ the determinant is different from zero. Thus, the columns form a basis and so it is possible to write the vector $\bf{b}$ as a linear combination of the columns.
Now, if $s=-3$ it is
$$\left(\begin{array}{rrr} -2 \\ 10 \\ -6 \end{array}\right)=4\left(\begin{array}{rrr} 1 \\ 4 \\ 0 \end{array}\right)-6\left(\begin{array}{rrr} 1 \\ 1 \\ 1 \end{array}\right).$$
Finally, if $s=3$ one has to look if there exist $\alpha$ and $\beta$ such that
$$\left(\begin{array}{rrr} 10 \\ 10 \\ 6 \end{array}\right)=\alpha\left(\begin{array}{rrr} 1 \\ 4 \\ 0 \end{array}\right)+\beta\left(\begin{array}{rrr} 1 \\ 1 \\ 1 \end{array}\right).$$
(Note that $(9,9,9)^T=9(1,1,1)$ and thus we can omit it). Show that there is no solution. So, the vector ${\bf b}$ cannot be written in terms of the columns. An alternative way to show this is to prove that 
$$\mathrm{det}\left(\begin{array}{rrr} 10 & 1 & 4\\ 10 & 4 & 1 \\ 6 & 0& 1 \end{array}\right)\ne 0.$$ In such a case the columns are linearly independent.
