Assignment: determining sets are bases of $\mathbb{R}^3$ This is question from an assignment I'm working on:

Which two of the following three sets in $\mathbb{R}^3$ is a basis of $\mathbb{R}^3$?
\begin{align*}
B_1&=\{(1,0,1),(6,4,5),(-4,-4,7)\}\\
B_2&=\{(2,1,3),(3,1,-3),(1,1,9)\}\\
B_3&=\{(3,-1,2),(5,1,1),(1,1,1)\}
\end{align*}

Thanks to software, I know that the answer is $B_1$ and $B_3$, as $B_2$ the only linearly dependent set of the three - and in this case, an LD set can't be a basis for $\mathbb{R}^3$.
Manually, I've put all three sets in RREF, and all three can be reduced to
$$
\left[\begin{array}{ccc|c}
{1}&{0}&{0}&{0}\\
{0}&{1}&{0}&{0}\\
{0}&{0}&{1}&{0}
\end{array}\right]
$$
This also checks out when computed by software.
Since all three sets reduce to the same RREF, how can I prove that these sets are linearly (in)dependent?
 A: Let V={$v_1,v_2,...,v_n$} be a set of n vectors in $\mathbb{R}^n$. then V is linearly independent iff the matrix A=($v_1,v_2,...,v_n$) is invertible where v_i is the ith column of A . proving this result is an easy exercise
. 
A: For $B_{2}$, this is linearly independent if the only solution to 
$$a(2,1,3)^{T} + b (3,1,-3)^{T} +c(1,1,9)^{T} = 0$$
has the trivial solution $a=b=c=0$
or 
$$
\begin{bmatrix}
2&3&1\\
1&1&1\\
3&-3&9
\end{bmatrix}\begin{bmatrix}
a\\
b\\
c
\end{bmatrix} = \begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$$
Take $-2R_{2} +R_{1}\rightarrow R_{1}$, $-3R_{2} + R_{3}\rightarrow R_{3}$ $-1/6 R_{3}\rightarrow R_{3}$
To get 
$$
\begin{bmatrix}
0&1&-1\\
1&1&1\\
0&1&-1
\end{bmatrix}\begin{bmatrix}
a\\
b\\
c
\end{bmatrix} = \begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$$
So the rank is clearly less than 3 hence there are infinitely many solutions. If we continue by taking $-1R_{1} + R_{3}\rightarrow R_{3}$ and perform a row swap, we get
$$
\begin{bmatrix}
1&1&1\\
0&1&-1\\
0&0&0
\end{bmatrix}\begin{bmatrix}
a\\
b\\
c
\end{bmatrix} = \begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$$
At this point, you can either continue taking the system to RREF, note that since the system has a row of zeros and is homogeneous, it must be infinitely many solutions, or perform back-substitution beginning with $c=s$ where $s\in \mathbb{R}$.
The choice of row operations are more or less arbitrary, I mainly did them because of the $1$ in row 2 for ease of calculations. It isn't precisely Gaussian Elimination because I did not start with a row exchange so that the leading 1 is in the first pivot column, but that does not matter. As soon as we got a row of zeros, that showed that Row 3 was a linear combination of Row 1 and Row 2, hence the dimension that these three vectors spans is $2$ so it cannot span $\mathbb{R}^{3}$.
