I have a question that after reduced echelon form we get an equation .if we have k1,k2,k3 all zero then what would be the solution? Is it unique solution or infinite no of solution? And what will we call it ? Linearly dependent or independent. ?
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$\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. $\endgroup$– seldonMar 9, 2016 at 18:49
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1$\begingroup$ I have this question. Which of the following sets of vectors are linearly dependent? One of the set is (5,2,0,-1), (0,-3,0,1), (1,0,-1,2), (3,1,0,1). How to solve this ? $\endgroup$– user317685Mar 9, 2016 at 19:02
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$\begingroup$ Change your question to that and add what you've tried $\endgroup$– seldonMar 9, 2016 at 19:08
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$\begingroup$ I tried this by reduced echelon form and system of equation has solutions like k=1, l=2 , m=0 so this type of solution what we call it ? $\endgroup$– user317685Mar 9, 2016 at 19:11
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Welcome to stack exchange. If you could at least post your reduced echelon matrix it would be helpful so we could try and help you
\begin{bmatrix}5&2&0&-1\\0&-3&0&1\\1&0&-1&2\\3&1&0&1\end{bmatrix}
\begin{bmatrix}1&0&-1&2\\0&-3&0&1\\5&2&0&-1\\3&1&0&1\end{bmatrix}
\begin{bmatrix}1&0&-1&2\\0&-3&0&1\\0&2&5&-11\\0&1&3&-5\end{bmatrix}
\begin{bmatrix}1&0&-1&2\\0&1&3&-5\\0&2&5&-11\\0&-3&0&1\end{bmatrix}
\begin{bmatrix}1&0&-1&2\\0&1&3&-5\\0&0&-1&-1\\0&0&9&-14\end{bmatrix}
Can you understand the steps I have taking?. Now can you say if the vectors are linearly independent or dependent
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$\begingroup$ I m new here so i don't know how to write or post my question here ? $\endgroup$ Mar 9, 2016 at 19:16
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$\begingroup$ This should help you meta.math.stackexchange.com/questions/12777/… $\endgroup$– EamonnMar 9, 2016 at 19:17
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$\begingroup$ If you cant understand it just post your solution and we will format it for you. Just type lets say 1 1 1 2 newline 2 2 3 4 newline 1 3 4 5. $\endgroup$– EamonnMar 9, 2016 at 19:18
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$\begingroup$ 1 0 0 0 newline 0 1 2 2 newline 0 0 1 0 newline 0 0 0 0 $\endgroup$ Mar 9, 2016 at 19:21