Hence it is shown because there no complimentary variable it is straight independent. (I think)
My question is is it likewise straight indepedent because the vectors are not multiples of the first vector.
You are watching: Determine if the columns of the matrix form a linearly independent set. justify your answer.
edited Sep 29 "14 at 19:54
Fernancarry out Martinez
asked Sep 19 "14 at 17:40
Fernando MartinezFernanexecute Martinez
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To answer your question
My question is is it likewise straight indepedent because the vectors are not multiples of the initially vector.
No. Just bereason the second and third columns are not multiples of the initially, it does not expect they are lialmost independent. Take for instance the matrix
$$eginpmatrix 1 & 1 & 1\1&2&3endpmatrix$$
Namong the columns are multiples of the others, however the columns carry out form a lialmost dependent collection. You know this without any genuine work, because $3$ vectors in $carolannpeacock.combbR^2$ cannot form a livirtually independent collection.
answered Sep 19 "14 at 17:51
David PDavid P
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You are appropriate that after row reducing and also finding that tbelow are no totally free variables (bereason eextremely column has a pivot), then all of the columns are lipractically independent.
By knowing the collection of three vectors is linearly independent, we recognize that the third column vector cannot be composed as a straight combination of the first column vector and also the second column vector. That is, tbelow carry out not exist $c_1$, $c_2 in carolannpeacock.combbR$ such that $v_3 = c_1v_1 + c_2v_2$. ($v_i$ is my notation for the $i$-th column vector.)
Similarly, we recognize $v_2$ cannot be written as a straight combicountry of $v_1$ and also $v_3$.
We likewise know $v_1$ cannot be created as a linear combicountry of $v_2$ and $v_3$.
That is what is supposed by the $3$ vectors being livirtually independent. You can not create any one of them as a linear combination of the others.
answered Sep 19 "14 at 17:52
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