I wanted to compute the span for the following sets in the vector space $\Bbb R^3$ over the field $\Bbb R$
$\{(1,6,9),(1,2,4),(5,22,34)\}$ and $\{(1,6,9),(1,2,4),(5,22,35)\}$
I began setting up the system of linear equations as below. $a(1,6,9)+b(1,2,4)+c(5,22,34)=(x,y,z)$ which is in the span.
Thus $x=a+b+5c$, $y=6a+2b+22c$, $z=9a+4b+34c$
Upon solving these for a, b, c, I get that each one is in terms of $x, y$, and $z$.
My first question is: Why does this mean that the span is $\Bbb R^3$? Would it not be $\Bbb R^3$ if a was in terms of just $x$, or $x$ and $y$?
Second question: for $(1,6,9),(1,2,4),(5,22,35)$, I get that the span is the set $(x, 0, 1.5x)$. I have never done these before so I wanted to see if this was correct?
Thanks in advance!
NOTE: We are not allowed to use the determinant and must only use systems of linear equations to solve.
$$\text{span}\{\begin{pmatrix}1\\6\\9\end{pmatrix},\begin{pmatrix}1\\2\\4\end{pmatrix},\begin{pmatrix}5\\22\\34\end{pmatrix}\}=\Bbb R^3\text{ if every vectors in }\Bbb R^3\text{ is in the span}$$ That is to say, can we solve for $$a_1\begin{pmatrix}1\\6\\9\end{pmatrix}+a_2\begin{pmatrix}1\\2\\4\end{pmatrix}+a_3\begin{pmatrix}5\\22\\34\end{pmatrix}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$$ for every $x=(x_1.x_2,x_3)$, in fact we can always find such $a_1,a_2,a_3$ because the above equation can be resolved into $$\begin{pmatrix}1&1&5\\6&2&22\\9&4&34\end{pmatrix}\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$$ And $\det\begin{pmatrix}1&1&5\\6&2&22\\9&4&34\end{pmatrix}\neq0 $ ensure that we can find such $(a_1,a_2,a_3)$
For the second set, we cannot find such $(a_1,a_2,a_3)$ for every $x\in \Bbb R^3$, because the determinant equals $0$ in the second case. Also, $$3\begin{pmatrix}1\\6\\9\end{pmatrix}+2\begin{pmatrix}1\\2\\4\end{pmatrix}=\begin{pmatrix}5\\22\\35 \end{pmatrix}$$ So the span of the second set is actually the span of the first two vectors, in which the first two are linearly independent to each other, then the span of these two vectors is essentially a plane passing through origin and the two vectors. What left is to find an equation of the plane passing through $O$ and $(1,6,9)$ and $(1,2,4)$