1. 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.

2. Re: Two quick questions

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

3. Re: Two quick questions

$\begin{pmatrix}1 & -4 & -2 & 0 & 3 & - 5 \\ 0 & 0 & 1 & 0 & 2 & -1 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$
is equivalent to the three equations
$x_1- 4x_2- 2x_3+ 3x_5- 5x_6= 0$
$x_3+ 2x_5- x_6= 0$
$x_6= 0$

From the last, $x_6= 0$ so that $x_3+ 2x_5= 0$ and $x_3= -2x_5$. Putting that into the first equation, $x_1- 4x_2- 2(-2x_5)+ 3x_5= x_1- 4x_2+ 7x_5= 0$. Then the best we can do is say $x_1= 4x_2- 7x_5$ and take $x_2= r$, $x_4= s$ and $x_5= t$ as parameters, $x_1= 4r- 7t$, $x_2= r$, $x_3= -2t$, $x_4= s$, $x_5= t$, $x_6= 0$.