Explain why $S$ is not a basis for $\mathbb{R}^3$ Explain why $S$ is not a basis for $\mathbb{R}^3$
$S=\{(1, 3, 0),(4, 1, 2),(-2, 5, -2)\}$
I set this equal to an arbitrary vector $\mathbf{x} = (x_1, x_2, x_3)$
After solving I got the matrix:
$\begin{bmatrix} 1 & 4 & -2 \\ 3 & 1 & 5 \\ 0 & 2 & -2\end{bmatrix}$
And since the $det(A)=0$ this set does not span $\mathbb{R}^3$.  However, I don't understand the answer the solution gave. It says:

A basis for $\mathbb{R}^3$ contains three linearly independent vectors. Because $-2(1, 3, 0)+(4, 1, 2)+(-2, 5, -2)=(0, 0, 0)$
$S$ is linearly dependent and is, therefore, not a basis for $\mathbb{R}^3$

However, if I test for linear independence I get the following $rref(A)$:
$\begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$
Which would imply to me that $S$ is linearly independent given the $0$ column.
 A: Your RREF tells you that $S$ is linearly independent.  What you have done is row-reduced the augmented matrix:
$$\left(\begin{array}{ccc|c}
1 & 4 & -2 & 0\\
3 & 1 & 5 & 0\\
0 & 2 & -2 & 0
\end{array}\right)$$
to the reduced row echelon form
$$\left(\begin{array}{ccc|c}
1 & 0 & 2 & 0\\
0 & 1 & -1 & 0\\
0 & 0 & 0 & 0
\end{array}\right)$$
What this tells you is that the matrix formed by setting the columns equal to the given vectors of $S$ has non-trivial nullspace, meaning there exist constants $c_1,c_2,c_3$ not all $0$, such that $c_1 (1,3,0) + c_2(4,1,2) + c_3(-2,5,-2) = (0,0,0)$
What we will have done is effectively checked whether the 3 vectors given satisfy the definition of linear independence.  Since we have nonzero constants $c_1,c_2,c_3$ so that $c_1 (1,3,0) + c_2(4,1,2) + c_3(-2,5,-2) = (0,0,0)$, they are not linearly independent.
In fact, by simply backsolving and picking $c_3 = 1$, we could choose $c_2 = 1$, $c_1 = -2$
That implies that $\{(1,3,0),(4,1,2),(-2,5,-2)\}$ are linearly dependent, as the solution in the book states.  And 3 vectors in $\mathbb{R}^3$ are a basis of $\mathbb{R}^3$ if and only if they are linearly independent, so the 3 vectors given are not a basis of $\mathbb{R}^3$.

I previously said that this is because you have a row of zeros.  Technically, this is wrong.  The correct explanation has to do with pivot columns, namely that we have a non-leading non-pivot column on the LHS when reducing to row echelon form.  A row of zeros is enough if you are dealing with a square matrix, but seeing as that is rarely the case, you will want to know the more general method.
Say you have $4$ vectors in $\mathbb{R}^6$.  These are linearly independent if and only if the matrix formed with the vectors has rank $4$.  Another way to verify this is that if each and every column on the LHS of the augmented matrix (when reduced to row echelon form) is a pivot column, then the column vectors forming your matrix are linearly independent.
A: Because
$$ 2(1,3,0) - (4,1,2) = (-2,5,-2) $$
A: This is an amplification of Lemur's answer, expanding on the "see your vectors and play with them" in his comment.  We want to know if we can find scalars $\alpha,\beta,\gamma$ such that
$$\alpha(1,3,0)+\beta(4,1,2)+\gamma(-2,5,-2)=(0,0,0)\ .$$
But looking at the third component gives immediately $\beta=\gamma$.  So the previous equation can be written as
$$\alpha(1,3,0)+\beta(2,6,0)=(0,0,0)\ ,$$
and now it is easy to see that $\alpha=2$, $\beta=-1$ is a solution.  So
$$2(1,3,0)-(4,1,2)-(-2,5,-2)=(0,0,0)\ ,$$
which shows that the vectors are linearly dependent.
