Thus the is shown because there no cost-free variable that is straight independent. (I think)

My concern is is it additionally linear indepedent due to the fact that the vectors room not multiples the the very first vector.

You are watching: Determine if the columns of the matrix form a linearly independent set. justify your answer.

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edited Sep 29 "14 at 19:54
Fernando Martinez
request Sep 19 "14 at 17:40

Fernando MartinezFernando Martinez
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My inquiry is is it also linear indepedent since the vectors room not multiples of the first vector.

No. Just due to the fact that the 2nd and third columns room not multiples of the first, that does not typical they space linearly independent. Take it for instance the matrix

$$\beginpmatrix 1 & 1 & 1\\1&2&3\endpmatrix$$

None that the columns room multiples that the others, however the columns do kind a linearly dependency set. You know this without any type of real work, due to the fact that $3$ vectors in $\carolannpeacock.combbR^2$ cannot form a linearly elevation set.

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reply Sep 19 "14 in ~ 17:51

David PDavid ns
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You are right that after row reducing and finding the there are no totally free variables (because every column has a pivot), then all of the columns space linearly independent.

By knowing the set of three vectors is linearly independent, we understand that the 3rd column vector cannot be composed as a linear mix of the an initial column vector and the second column vector. That is, there do not exist $c_1$, $c_2 \in \carolannpeacock.combbR$ such the $v_3 = c_1v_1 + c_2v_2$. ($v_i$ is mine notation because that the $i$-th pillar vector.)

Similarly, we recognize $v_2$ cannot be composed as a linear combination of $v_1$ and $v_3$.

We likewise know $v_1$ can not be created as a linear combination of $v_2$ and $v_3$.

That is what is intended by the $3$ vectors being linearly independent. You can"t write any type of one of them as a linear mix of the others.

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answered Sep 19 "14 at 17:52

laymanlayman
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