Question about an example procedure to obtain a basis from L.I list in Axler's Linear algebra In the Bases section of the book there is an example that says:
As an example in $F^3$ suppose we start with the linearly independent list $\{(2,3,4),(9,6,8)\ \} \rightarrow (\alpha)$. If we take
$w_1,w_2,w_3$ in the proof to be the standard basis of $F^3$ then the procedure produces the list $(2,3,4),(9,6,8),(0,1,0)$ which is a basis of $F^3$
Step 1
If $v_1=0$, delete it from B. If $v_1 \neq 0 $, leave B unchanged.
Step j
If $v_j$ is in $span(v_1,...,v_{j-1})$ delete $v_j$ from B. If $v_j$ is not in the span, leave B unchanged.
Applying the procedure above to the list $u_1,...,u_m,w_1,...,w_n$  (u's are l.i and w's are basis of $V$) to reduce this list to a basis of V produces a list of the vectors $u_1,...u_m$ (none of the us get deleted because they are L.I) and some of the w's.
The problem is that, I do not know where the vector $\mathbf{(0,1,0)}$ came from (I cannot assume using it directly in the definitions, there must be an origin since proposing immediately $(1,0,0)$ or $(0,0,1)$ may be a waste of time).
So I know that ($\alpha$) is linearly independent and the steps above tells us that a vector $v_j$ might be in the $span(v_1,...,v_{j-1})$.
In the section of span, the set of all linear combinations of a list of vectors $v_{1},...,v_{m}$ in $V$ is called the span of $v_{1},...,v_{m}$.
Then, I proceeded to solve the system:
(Taking $(9,6,8)$ as the $v_{jth}$ vector)
$(9,6,8)=a_1(2,3,4)+a_2(9,6,8)$
Resulting in:
$a_1=0$ and  $a_2=1$,
and taking $a_3=0$ generates the vector $(0,1,0)$
The procedure tells us not to delete the rest of the vectors in $\alpha$, because they are L.I, and now with the added vector it fulfills the definition of a basis.
So my final question is:
Is my procedure (reasoning) right?
Is there another way to obtain that vector?
 A: There's a much more efficient way. Consider the vectors as columns; you add to the given linearly independent vectors a known basis of the space (whatever you like, not necessarily the standard one). With the standard basis you get
\begin{bmatrix}
2 & 9 & 1 & 0 & 0 \\
3 & 6 & 0 & 1 & 0 \\
4 & 8 & 0 & 0 & 1
\end{bmatrix}
and perform Gaussian elimination:
\begin{align}
\begin{bmatrix}
2 & 9 & 1 & 0 & 0 \\
3 & 6 & 0 & 1 & 0 \\
4 & 8 & 0 & 0 & 1
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 9/2 & 1/2 & 0 & 0 \\
3 & 6 & 0 & 1 & 0 \\
4 & 8 & 0 & 0 & 1
\end{bmatrix}
&& R_1\gets \tfrac{1}{2}R_1 \\[6px]
&\to
\begin{bmatrix}
1 & 9/2 & 1/2 & 0 & 0 \\
0 & -15/2 & -3/2 & 1 & 0 \\
0 & -10 & -2 & 0 & 1
\end{bmatrix}
&& \begin{aligned}R_2&\gets R_2-3R_1 \\ R_3&\gets R_3-4R_1\end{aligned}\\[6px]
&\to
\begin{bmatrix}
1 & 9/2 & 1/2 & 0 & 0 \\
0 & 1 & 1/5 & -2/15 & 0 \\
0 & -10 & -2 & 0 & 1
\end{bmatrix}
&& R_2\gets-\tfrac{2}{15}R_2 \\[6px]
&\to
\begin{bmatrix}
1 & 9/2 & 1/2 & 0 & 0 \\
0 & 1 & 1/5 & -2/15 & 0 \\
0 & 0 & 0 & -4/3 & 1
\end{bmatrix}
&& R_3\gets R_3+10R_2
\end{align}
You see that the fourth column becomes the third pivot column. And you knew at the outset that the matrix has rank $3$.
This means that columns 1, 2 and 4 are linearly independent in all the previously written matrices, because elementary row operations don't change the linear relation among columns. In particular, the first two columns will necessarily be pivot column in the elimination.
We conclude that columns 1, 2 and 4 in the initial matrix are linearly independent.
No need to test the vector in the “adjoined” basis one by one, because you can do it at once.
