
Two quick questions
Question 1
If I have matrix v1 = (1,1,2) v2 = (4,6,7) and v3 (1,1,k)
Am I right in saying that if k = 2, v1,v2,v3 is linearly dependent, because v3 = 1(v1)?
And following that, if I had to write one vector as a linear combination of the other two, would it be correct to say v1 = 0(v2)  1(v3)
Question 2
The following matrix is row equivalent to A. Find the general solution to Ax = 0
1  4  2  0  3  5 
0  0  1  0  2  1 
0  0  0  0  0  1 
0  0  0  0  0  0 
I get that the basic variables are x1 and x3, since they are pivots. But is x6 a basic variable or is it free? I'm confused because 1x6 = 0 and 0x6 = 0.

Re: Two quick questions
Question 1: yes
Question 2:
What you end up getting after bringing that matrix on RREF is
$\displaystyle A=\begin{pmatrix}1&4&0&0&7&0\\ 0&0&1&0&2&0\\ 0&0&0&0&0&1\\ 0&0&0&0&0&0 \end{pmatrix}$,
we see that the free variables are $\displaystyle x_2,x_4,x_5 $ and the lead variables are $\displaystyle x_1,x_3,x_6 $ since they are the pivots.
You ask if $\displaystyle x_6 $ are free, it's not because the only value it can take is 0 since it is a pivot in the third row.

Re: Two quick questions
$\displaystyle \begin{pmatrix}1 & 4 & 2 & 0 & 3 &  5 \\ 0 & 0 & 1 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$
is equivalent to the three equations
$\displaystyle x_1 4x_2 2x_3+ 3x_5 5x_6= 0$
$\displaystyle x_3+ 2x_5 x_6= 0$
$\displaystyle x_6= 0$
From the last, $\displaystyle x_6= 0$ so that $\displaystyle x_3+ 2x_5= 0$ and $\displaystyle x_3= 2x_5$. Putting that into the first equation, $\displaystyle x_1 4x_2 2(2x_5)+ 3x_5= x_1 4x_2+ 7x_5= 0$. Then the best we can do is say $\displaystyle x_1= 4x_2 7x_5$ and take $\displaystyle x_2= r$, $\displaystyle x_4= s$ and $\displaystyle x_5= t$ as parameters, $\displaystyle x_1= 4r 7t$, $\displaystyle x_2= r$, $\displaystyle x_3= 2t$, $\displaystyle x_4= s$, $\displaystyle x_5= t$, $\displaystyle x_6= 0$.