Thus the is shown since there no totally free variable it is straight independent. (I think)

My question is is it also linear indepedent because the vectors space not multiples that the very first vector.

You are watching: Determine if the columns of the matrix form a linearly independent set

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edited Sep 29 "14 in ~ 19:54
Fernando Martinez
asked Sep 19 "14 in ~ 17:40

Fernando MartinezFernando Martinez
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My concern is is it likewise linear indepedent since the vectors are not multiples that the first vector.

No. Just since the second and 3rd columns space not multiples the the first, that does not mean they space linearly independent. Take for instance the matrix

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

None the the columns space multiples of the others, yet the columns do kind a linearly dependency set. You understand this without any real work, due to the fact that $3$ vectors in $\inter-base.netbbR^2$ cannot form a linearly elevation set.

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answered Sep 19 "14 at 17:51

David PDavid ns
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You are ideal that after heat reducing and also finding the there room no free variables (because every column has actually a pivot), then every one of the columns are linearly independent.

By knowing the collection of three vectors is linearly independent, we understand that the 3rd column vector cannot be composed as a linear combination of the very first column vector and also the second column vector. That is, over there do not exist $c_1$, $c_2 \in \inter-base.netbbR$ such the $v_3 = c_1v_1 + c_2v_2$. ($v_i$ is mine notation because that the $i$-th tower vector.)

Similarly, we recognize $v_2$ can not be created 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 also $v_3$.

That is what is supposed by the $3$ vectors being linearly independent. You can"t write any kind of one of them together a linear combination of the others.

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answered Sep 19 "14 in ~ 17:52

laymanlayman
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