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How can it be seen if the following matrix is linear dependent?

Let $A= \begin{bmatrix} 0 & -3 & 9& \\ 2&1& 7 \\ -1& 4 &-5 \\ 1&-4&-2 \end{bmatrix} $

First operation I perfomed was switch r1 to r4 and -2r1+r2

$A= \begin{bmatrix} 1 & -4 & -2& \\ 0&9& -3 \\ -1& 4 &-5 \\ 0&-3&9 \end{bmatrix} $

Next performed was $r1+r3$ to get

Let $A= \begin{bmatrix} 1& -4 & -2& \\ 0&9& -3\\ 0& 0 &-7 \\ 0&-3&9 \end{bmatrix} $

next done was$-1/3r2+r4$ and got

Let $A= \begin{bmatrix} 1& -4 & -2& \\ 0&9& -3\\ 0& 0 &-7 \\ 0&0&-8 \end{bmatrix} $

Thus it is shown because there no free variable it is linear independent. (I think)

My question is is it also linear indepedent because the vectors are not multiples of the first vector.

Fernando Martinez
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  • Hint. Look at the $3$rd and $4$th vector... – the_candyman Sep 19 '14 at 17:44
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    I think you should be extra clear when you say something like "next performed $r_{1} + r_{3}$". Which row did this replace? $r_{1}$ or $r_{3}$? I know it can be determined by looking at the matrix, but for the sake of clarity, I think the detail should be added in the future. You might write instead "next performed $r_{1} = r_{1} + r_{3}$" if you replaced $r_{1}$ with $r_{1} + r_{3}$. – layman Sep 19 '14 at 17:44
  • Ok I shall do that right now... – Fernando Martinez Sep 19 '14 at 17:51

2 Answers2

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To answer your question

My question is is it also linear indepedent because the vectors are not multiples of the first vector.

No. Just because the second and third columns are not multiples of the first, it does not mean they are linearly independent. Take for example the matrix

$$\begin{pmatrix} 1 & 1 & 1\\1&2&3\end{pmatrix}$$

None of the columns are multiples of the others, but the columns do form a linearly dependent set. You know this without any real work, since $3$ vectors in $\mathbb{R}^2$ cannot form a linearly independent set.

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

By knowing the set of three vectors is linearly independent, we know that the third column vector cannot be written as a linear combination of the first column vector and the second column vector. That is, there do not exist $c_{1}$, $c_{2} \in \mathbb{R}$ such that $v_{3} = c_{1}v_{1} + c_{2}v_{2}$. ($v_{i}$ is my notation for the $i$-th column vector.)

Similarly, we know $v_{2}$ cannot be written as a linear combination of $v_{1}$ and $v_{3}$.

We also know $v_{1}$ cannot be written as a linear combination of $v_{2}$ and $v_{3}$.

That is what is meant by the $3$ vectors being linearly independent. You can't write any one of them as a linear combination of the others.

layman
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