Extending the question: Is $S$ a basis for $P^3$? 
Let $S = \{4 - t, t^3, 6t^2, 3t + t^3, -1 + 4t\}$.  Is $S$ a basis for $P^3$?

Polynomial Equation:
0t3 + 0t2 - t + 4 = 0
t3 + 0t2 + 0t + 0 = 0
0t3 + 6t2 + 0t + 0 = 0
t3 + 0t2 + 3t + 0 = 0
0t3 + 0t2 + 4t - 1 = 0  
Transform into an augmented matrix:
\begin{bmatrix}0&0&-1&4&0\\1&0&0&0&0\\0&6&0&0&0\\1&0&3&0&0\\0&0&4&-1&0\end{bmatrix}
RREF:
\begin{bmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&0\end{bmatrix}
I answered the question saying that S is not a basis for p3 because the all zero bottom row means that there is a free variable within the equation, making the overall system linearly dependent and therefore not a basis.
To push my understanding, can I further say that the vector (-1 + 4t) is the only dependent variable and if it was not in S, then S would be a basis of p3? 
Also, the presence of the free variable would mean that the dependent vector would be a non-unique combination of ANY of the other vectors found within S?

$P^3$ is 4 dimensional with a natural basis, $1, t, t^2, t^3$. In this basis, we can represent $S$ as the set of columns of
\begin{pmatrix}
 4 & 0 & 0 & 0 & -1 \\
 -1 & 0 & 0 & 3 & 4 \\
 0 & 0 & 6 & 0 & 0 \\
 0 & 1 & 0 & 1 & 0 \\
\end{pmatrix}
This row reduces to
\begin{pmatrix}
 1 & 0 & 0 & 0 & -\frac{1}{4} \\
 0 & 1 & 0 & 0 & -\frac{5}{4} \\
 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 1 & \frac{5}{4} \\
\end{pmatrix}
Row reducing gives us a basis for the relations between columns (the null space). Specifically, we have one relation
$$ (4 - t) + 5 (t^3) - 5(3t + t^3) + 4(-1 + 4t) = 0 $$
I can solve this equation for any element of $S$ except $6t^2$. This means that when I remove $4-t$ or $t^3$ or $3t+t^3$ or $-1 + 4t$ from $S$, I get a basis (because I no longer have the above relation and because I have a way of writing the removed vector in terms of the remaining vectors [using the above relation]).
