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I put the vectors in a matrix and reduced it, solved the determinant and got 0. This tells me that the vectors are linearly dependent. I am not sure how to figure out the non-trivial relation. This is my reduced matrix.


$$ \begin{pmatrix} 1& 0&-1/4 \\ 0& 1& 1\\ 0& 0& 0 \end{pmatrix} $$


$$ A = \left\{\begin{matrix} -60\\ -4\\ -72 \end{matrix}\right. : B = \left\{\begin{matrix} -5\\ -1\\ -5 \end{matrix}\right. : C = \left\{\begin{matrix} 10\\ 0\\ 13 \end{matrix}\right. $$

__A +__B +___C = 0

Find coefficients.

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  • $\begingroup$ Your reduced matrix corresponds to the system of equations, $a+b-(1/4)c=0,b+c=0$. Find a non-zero solution to that system. That solution will be the coefficients in the relation. $\endgroup$ – Gerry Myerson Feb 20 '14 at 8:42
  • $\begingroup$ Since they're dependent, can I set A = 1? $\endgroup$ – KnowledgeGeek Feb 20 '14 at 8:47
  • $\begingroup$ Also, shouldn't your equation just read a - (1/4)c? $\endgroup$ – KnowledgeGeek Feb 20 '14 at 8:48
  • $\begingroup$ How can you set $A=1$, when $A=\pmatrix{-60\cr-4\cr-72\cr}$? Or do you mean $a$ when you write $A$? In that case, set $a=1$, and see what happens. $\endgroup$ – Gerry Myerson Feb 20 '14 at 8:49
  • $\begingroup$ My 1st equation comes from the first row. The first row is $(1,1,-1/4)$. So, my 1st equation is $a+b-(1/4)c=0$. $\endgroup$ – Gerry Myerson Feb 20 '14 at 8:50
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Found the RREF, set the free variable to 1, since it's linearly dependent and solved the system of equations to get coefficients.

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