anonymous
  • anonymous
Are the following set of vectors in R^4 linearly independent? \[\left\{ (1,0,2,-1),(1,2,-1,0),(1,1,0,-1) \right\}\] I want to get the set of vectors into matrixform but the "set" term confuses me a bit. Is this the right form? \[\left[\begin{matrix}1 & 0 &2&-1 \\1 & 2 &-1&0 \\1 & 1&0&-1\end{matrix}\right]=\left[\begin{matrix}0 \\ 0 \\0\end{matrix}\right]\] Or is it the other way around?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
you need to add a row o zeroes
anonymous
  • anonymous
Add a row of zeroes, what do you mean by that?
anonymous
  • anonymous
|dw:1359278439800:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
That's how i ussualy write it, is that what you meant?
amoodarya
  • amoodarya
a(1,0,2,−1)+b(1,2,−1,0)+c(1,1,0,−1)=(0,0,0,0) they are independent if only a=b=c=0 multiply them and solve the system of equation
amoodarya
  • amoodarya
a+b+c=0 0a+2b+c=0 2a-b+0c=0 -1a+0b-c=0
anonymous
  • anonymous
\[a _{1}\left(\begin{matrix}1 \\0 \\ 2\\-1 \end{matrix}\right)+a _{2}\left(\begin{matrix}1 \\2 \\ -1\\0\end{matrix}\right)+a _{3}\left(\begin{matrix}1 \\1 \\ 0\\-1\end{matrix}\right)=\left(\begin{matrix}0 \\0 \\ 0\\0\end{matrix}\right)\] \[\left[\begin{matrix}1 & 1 & 1 \\0 & 2 & 1 \\ 2 & -1 & 0 \\-1 & 0 & -1 \end{matrix}\right]=\left(\begin{matrix}0 \\ 0 \\ 0 \\0\end{matrix}\right)\] \[\left[\begin{matrix}1 & 0 &0 \\0& 1 & 0 \\ 0&0&1 \\ 0&0&0\end{matrix}\right]=\left[\begin{matrix}0 \\ 0 \\0\\0\end{matrix}\right]\]
anonymous
  • anonymous
@amoodarya How should i interpret the reduced matrix, 000=0 but also 001=0 ?
anonymous
  • anonymous
So it can't really be a_3=t it must be a_3=0 and a_1=a_2=a_3=0 and the vectors are linearly independent, am I correct?
amoodarya
  • amoodarya
do you "rank of matrix" ?
anonymous
  • anonymous
Don't really know what rank of matrix is, will have to look it up.
anonymous
  • anonymous
@amoodarya So the number of linearly independent vectors in the basis is three, so the rank should be three too, but what does this say about the dependent/independent case?

Looking for something else?

Not the answer you are looking for? Search for more explanations.