1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.