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anonymous
 3 years ago
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?
anonymous
 3 years ago
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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you need to add a row o zeroes

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Add a row of zeroes, what do you mean by that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359278439800:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's how i ussualy write it, is that what you meant?

amoodarya
 3 years ago
Best ResponseYou've already chosen the best response.1a(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
 3 years ago
Best ResponseYou've already chosen the best response.1a+b+c=0 0a+2b+c=0 2ab+0c=0 1a+0bc=0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[a _{1}\left(\begin{matrix}1 \\0 \\ 2\\1 \end{matrix}\right)+a _{2}\left(\begin{matrix}1 \\2 \\ 1\\0\end{matrix}\right)+a _{3}\left(\begin{matrix}1 \\1 \\ 0\\1\end{matrix}\right)=\left(\begin{matrix}0 \\0 \\ 0\\0\end{matrix}\right)\] \[\left[\begin{matrix}1 & 1 & 1 \\0 & 2 & 1 \\ 2 & 1 & 0 \\1 & 0 & 1 \end{matrix}\right]=\left(\begin{matrix}0 \\ 0 \\ 0 \\0\end{matrix}\right)\] \[\left[\begin{matrix}1 & 0 &0 \\0& 1 & 0 \\ 0&0&1 \\ 0&0&0\end{matrix}\right]=\left[\begin{matrix}0 \\ 0 \\0\\0\end{matrix}\right]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@amoodarya How should i interpret the reduced matrix, 000=0 but also 001=0 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.1do you "rank of matrix" ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Don't really know what rank of matrix is, will have to look it up.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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?
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