MHB Are These Vectors Linearly Dependent?

Click For Summary
The vectors provided are linearly dependent, as demonstrated through the row reduction of their corresponding matrix. By solving the equations derived from the linear combination of the vectors, it was established that there exist non-zero coefficients satisfying the dependency condition. The discussion also highlights the importance of understanding the definitions of linear independence and dependence. Participants reflected on their methods, questioning the necessity of row reduction if the dependency was already suspected. Overall, the conclusion confirms that the vectors are indeed linearly dependent.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
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
 
Last edited:
Physics news on Phys.org
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?
 
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
https://www.physicsforums.com/attachments/9045
 
Last edited:
I edited the thread title. The original title of "12.1" wasn't of much use to describe the topic. :)
 
Thank you

I tried also to change it

But when I submitted it didn't happen
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K