To express a vector as a linear combination of vectors, do the vectors need to linearly independent? For example, write (-3,4,1) as a linear combination of (1,2,1) (8,1,2) and (4,3,2)
Is this possible considering these vectors are not linearly independent?
I ask this because when constructing these into a matrix and running a reduced row echelon form, I do not have a clear answer, and I figure this is the reason.
Thank you for the help.
 A: In order to determine whether it is possible to write $(-3,4,1)$ as a linear combination of $(1,2,1), (8,1,2),$ and $(4,3,2),$ the standard approach is to put these vectors together as the columns of a matrix, then row reduce. In particular, we have
$$
\left(\begin{array}{cccc} 1 & 8 & 4 & -3\\ 2 & 1 & 3 & 4\\ 1 & 2 & 2 & 1 \end{array}\right)
\leadsto \left(\begin{array}{cccc} 1 & 0 & \frac{4}{3} & \frac{7}{3}\\ 0 & 1 & \frac{1}{3} & -\frac{2}{3}\\ 0 & 0 & 0 & 0 \end{array}\right).
$$
Because the last column of the row-reduced matrix is not a pivot column, it is possible to write $(-3,4,1)$ as a linear combination of the other three vectors.  In particular: we see that in the row-reduced matrix, the last column can be written as $\frac 73$ of the first column added to $- \frac 23$ of the second column.  Because this holds for the row-reduced matrix, it also holds for the original matrix. That is, we have
$$
(-3,4,1) = \frac 73 \cdot (1,2,1) - \frac 23 \cdot (8,1,2).
$$
A: It doesn't have to be linearly independent. 
The RREF is $$\begin{bmatrix} 1 & 0 & \frac{4}{3} & \frac{7}{3}\\ 0 & 1 & \frac{1}{3} & -\frac{2}{3}\\ 0 & 0 & 0 & 0 \end{bmatrix}$$
The system is consistent. 
As noted, there are only two pivot columns and that means we have a free variable. Let $x_3=t$, then we have 
$x_1=\frac73-\frac43t$ and $x_2=-\frac23-\frac13t$.
Each possible $t$ gives you a possible linear combination.
A: Not exactly.
In your example, suppose the coeffients are a b and c.
Then we want to solve the equation:
$$
\begin{bmatrix}
1 & 8 & 4\\
2 & 1 & 3\\
1 & 2 & 2
\end{bmatrix}
\begin{bmatrix}
a\\b\\c
\end{bmatrix}=
\begin{bmatrix}
-3 \\4 \\1
\end{bmatrix}
$$
Because the base vectors are dependant, the equation has no solutions that can be expressed as $x=A^{-1}b$. 
