Linear independence and dependence $u = (1, 1, 0)$, $v = (1, 4, 1)$, $w = (r^2, 1, r^2)$, $b = (3+2r, 5+12r, 2r)$
a) For which values of $r$, is the set $(u,v,w)$ linearly independent? --> I got $r = \pm 0.5$ via ERO. Could someone just show me the matrix, so I know if I was on the right track?
b) For which values of $r$ is the vector $b$ a linear combination of $u,v,w$, and for which of these values of $r$ can be written as a linear combination in more than one way?  --> I got $r = 1$ or $-2$ but how to work out this part "or which of these values of $r$ can be written as a linear combination in more than one way?"
Thanks folks
 A: Let the matrix A be the $3\times3$ matrix whose columns are the vectors $u, v\text{ and } w$.  Form the augmented matrix, $[A | b]$ . 
$$\left[ \begin{array}{ccc|c}1&1&r^2&3+2r\\1&4&1&5+12r\\0&1&r^2&2r\end{array} \right]$$
Do elementary row operations to get the left side of the augmented matrix into row echelon form.
$$\left[ \begin{array}{ccc|c}1&1&r^2 &3+2r\\1&4&1&5+12r\\0&1&r^2&2r
\end{array} \right] \Rightarrow \left[ \begin{array}{ccc|c}1&0&0&3\\1&4&1&5+12r\\0&1&r^2&2r\end{array} \right] \Rightarrow \left[ \begin{array}{ccc|c}1&0&0&3\\0&4&1&2+12r\\0&1&r^2&2r\end{array} \right] \Rightarrow \left[ \begin{array}{ccc|c}1&0&0&3\\0&4&1&2+12r\\0&0&r^2-\frac{1}{4}&-r-\frac{1}{2}\end{array} \right]$$
The last row is the key to the whole question. 
For part a) if $r^2-\frac{1}{4}=0$, the the rows of $A$ are linearly dependent and so too are the columns of the original matrix, $A$. Thus, If $r = \pm \frac{1}{2}$, the the vectors $u, v, w$ are linearly dependent. This is easily verified since in this case, $v = 4w$. The vectors are independent for any other value of $r$. Namely the vectors $u, v\text{ and } w$ are independent for any $r$ such that  $r \ne \pm \frac{1}{2}$.
For part b) we again see that if $r \ne \pm \frac{1}{2}$, the matrix $A$ is full rank and the system $Ax = b$ has a unique solution for any $b$. If, however, $r = \pm \frac{1}{2}$ the system has either infinitely many solutions or no solutions. Notice that when we do back substitution the last row says $(r^2-\frac{1}{4})x_3 = -(r+\frac{1}{2})$. If we let  $r = \frac{1}{2}$, then we get $0\,x_3 = -1$ and the system of equations is inconsistent and there is no solution. If however, we let with $r = -\frac{1}{2}$, we get $0\,x_3 = 0$ and there are infinitely many solutions.
