Sorry for my newbish ways but i don't know how to write out everything here yet so bare with me.

The question asks to write some vector in $S$ as a linear combination of the others.

the vectors are:

$v1 = [0,0,0]$

$v2 = [-2,3,-4]$

$v3 = [4,-3,2]$

i got the reduced row echelon form which is

\begin{matrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{matrix}

So what do i do from here? Is $x_1$ a free variable or just 0 or is $x_2$ a free variable?

I honestly have no idea where to start due to all the $x_1$'s being 0.

  • $\begingroup$ Yep! v1 is = [0,0,0] $\endgroup$ – Newb18 Sep 23 '15 at 22:00
  • $\begingroup$ x1 is a free variable. x2 isn't. Try to write the solution set now. $\endgroup$ – Symeof Sep 23 '15 at 22:03
  • $\begingroup$ How about this: $[0,0,0] = 0[-2,3,-4]+0[4,-3,2]$. Thus $[0,0,0]$ is a linear combination of the other two vectors. $\endgroup$ – got it--thanks Sep 23 '15 at 22:48
  • $\begingroup$ But isnt it linearly dependent so that means it would be [1,0,0]? $\endgroup$ – Newb18 Sep 23 '15 at 22:57
  • 1
    $\begingroup$ What does $[1,0,0]$ mean? If the question is asking you to express one vector as a linear combination of the others, then that's what I did: $[0,0,0] = 0\cdot [-2,3,-4]+0\cdot [4,-3,2]$. You don't need to solve a matrix equation to it in this case, because it's easy to see. If you do solve the matrix equation, you'll likely end up with $x_1$ is free, $x_2=x_3=0$, in which case it's just telling you that $k\cdot [0,0,0]+0\cdot [-2,3,-4]+0\cdot [4,-3,2]=[0,0,0]$. But set $k=1$ and move the other two vectors to the other side and you'll have what I wrote. $\endgroup$ – got it--thanks Sep 24 '15 at 13:43

Let us name these vectors as X, Y and Z respectively where X is the zero vector. Since one of the given vectors is a zero vector, they are certainly dependent and one can be written as a linear combination of the two others. One way to do this is: kX=0Y+0Z where k is any real.


By applying elementary row operations on your reduced row echelon form you will have the matrix:

\begin{matrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix}

Note that the first row have no leading 1, this means that $x_1$ is a free variable. So you are correct.

Knowing $x_1$ a free variable, you can now solve for the solution set of your equations.

  • $\begingroup$ So is the answer simply [1,0,0]??? $\endgroup$ – Newb18 Sep 23 '15 at 22:55
  • $\begingroup$ Or just [0,0,0]=0[−2,3,−4]+0[4,−3,2]? $\endgroup$ – Newb18 Sep 24 '15 at 1:58

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