# Determining Linear Independence/Dependence & non-trivial solution.

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.

• 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. – Gerry Myerson Feb 20 '14 at 8:42
• Since they're dependent, can I set A = 1? – KnowledgeGeek Feb 20 '14 at 8:47
• Also, shouldn't your equation just read a - (1/4)c? – KnowledgeGeek Feb 20 '14 at 8:48
• 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. – Gerry Myerson Feb 20 '14 at 8:49
• My 1st equation comes from the first row. The first row is $(1,1,-1/4)$. So, my 1st equation is $a+b-(1/4)c=0$. – Gerry Myerson Feb 20 '14 at 8:50