# 12.1 - Linear dependence

#### karush

##### Well-known member
Are the vectors
$$\left[ \begin{array}{r} 2\\1\\-2 \end{array}\right] ,\quad \left[\begin{array}{r} 0\\2\\-2 \end{array}\right] ,\quad \left[\begin{array}{r} 2\\3\\-4 \end{array}\right]$$
linearly dependent or linearly independent?
$$\left[ \begin{array}{rrr|r} 2 & 0 & 2 & 0 \\ 1 & 2 & 3 & 0 \\ -2 & -2 & -4 & 0 \end{array} \right] \sim \left[ \begin{array}{rrr|r} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$$
I assume this is independent due to trivial answers

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#### HallsofIvy

##### Well-known member
MHB Math Helper
Re: 12.1

By the definition of "linearly dependent", these vectors are linearly dependent if and only if there exist three number, a, b, and c, not all 0, such that
$$a\begin{bmatrix}2 \\ 1 \\ -2 \end{bmatrix}+ b\begin{bmatrix}0 \\ 2 \\ -2 \end{bmatrix}+ c\begin{bmatrix}2 \\ 3 \\ -4\end{bmatrix}= \begin{bmatrix}2a+ 2c \\ a+ 2b+ 3c \\ -2a- 2b- 4c \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$.

That is, 2a+ 2c= 0, a+ 2b+ 3c= 0, -2a- 2b- 4c= 0. From 2a+ 2c= 0, c= -a so the last two equations are a+ 2b- 3a= 2b- 2a= 0 and -2a- 2b+ 4c= 2a- 2b= 0. Both of those give a= b. Any numbers a, b, and c, such that b= a, c= -a will work.

So, yes, these vectors are linearly dependent.

You should think about two questions. What definition of "linearly independent" and "linearly dependent" did you learn? And what was your purpose in row reducing a matrix having the vectors as columns if you had to ask if the vectors were linearly dependent when you finished?

#### karush

##### Well-known member
Re: 12.1

I pretty much just followed an example
But probably could solve some of these just by observation
they used augmented matrix but only did some alteration
View attachment 9045

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