Lin. Alg. - Is the set a linearly independent subset of R^3

In summary, the question asks whether the given set of vectors is a linearly independent subset of R^3. Through row reduction, it is found that the system does not have a unique solution, indicating that the set is not linearly independent. This is because there is at least one free variable, allowing for infinite solutions to be generated.
  • #1
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Homework Statement


Is {(1,4,-6), (1,5,8), (2,1,1), (0,1,0)} a linearly independent subset of R^3. Justify your answer

Homework Equations


The Attempt at a Solution

I asssumed

a(1,4,-6) + b(1,5,8) + c(2,1,1) + d(0,1,0) = 0

then i set up the system

a + b + 2c = 0
4a + 5b + c + d = 0
-6a + 8b + c = 0

My first step was to switch the 2nd row with the 3rd row:

a + b + 2c = 0
-6a + 8b + c = 0
4a + 5b + c + d = 0

then i replaced the second row with ( 6R1 + R2)
and replaced the third row with (-4R1 + R3)

my result is

a+ b + 2c = 0
14b + 13c = 0
b - 7c + d = 0

then i replaced thethird row with ( - 1/14 R2 + R3)

a + b + 2c = 0
14b + 13c = 0
-111/14 c + d = 0

on this step, it's looking closer to what Dick got, and is there supposed to be another manipulation with the rows?

I just solved for d = 111/14 c

then i just let c = 1, thus
c = 1
d = 111/14
b = -13/14
a = (13/14) - 2

but if i let c = 14

i get dick's answer

c = 14
d = 111
b = -13
a = -15are both results correct??

and if not, (meaning Dick's is the only correct solution), what is the next step in the algorithm to find c = 14?

thanks for the helpnevermind, upon further reading i found that

"In this case, the system does not have a unique solution, as it contains at least one free variable. The solution set can then be expressed parametrically (that is, in terms of the free variables, so that if values for the free variables are chosen, a solution will be generated)."

so there's no unique solution, since you can choose whatever you want your variable "c" to be.
 
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  • #2
are you sure you row-reduced the system? even if you didn't, you should have encountered in your class that for R^n, the maximum number of vectors that can be linearly independent is n, which is in this case 3
 
  • #3
a=-15, b=-13, c=14, d=111. Yes, you've missed some solutions, as proton predicted.
 
  • #4
thanks alot, I'm actually taking the linear algebra course, with an intro to linear algebra course at the same time, so I'm not really familiar with this topic yet

but when you mentioned row reduced system, i looked it up, and worked out the problem
before, i just tried random substitutions

My first step was to switch the 2nd row with the 3rd row:

a + b + 2c = 0
-6a + 8b + c = 0
4a + 5b + c + d = 0

then i replaced the second row with ( 6R1 + R2)
and replaced the third row with (-4R1 + R3)

my result is

a+ b + 2c = 0
14b + 13c = 0
b - 7c + d = 0

then i replaced thethird row with ( - 1/14 R2 + R3)

a + b + 2c = 0
14b + 13c = 0
-111/14 c + d = 0

on this step, it's looking closer to what Dick got, and is there supposed to be another manipulation with the rows?

I just solved for d = 111/14 c

then i just let c = 1, thus
c = 1
d = 111/14
b = -13/14
a = (13/14) - 2

but if i let c = 14

i get dick's answer

c = 14
d = 111
b = -13
a = -15are both results correct??

and if not, (meaning Dick's is the only correct solution), what is the next step in the algorithm to find c = 14?

thanks for the help
 
  • #5
nevermind, upon further reading i found that

"In this case, the system does not have a unique solution, as it contains at least one free variable. The solution set can then be expressed parametrically (that is, in terms of the free variables, so that if values for the free variables are chosen, a solution will be generated)."

so there's no unique solution, since you can choose whatever you want your variable "c" to be.
 
  • #6
You've got it. My solution was just a 'for instance'.
 

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations, vectors, matrices, and their properties. It is used to solve systems of linear equations and analyze geometric transformations.

2. What is a linearly independent subset?

A linearly independent subset is a set of vectors in which none of the vectors can be written as a linear combination of the other vectors. In other words, no vector in the subset is redundant and each vector contributes to the overall span of the set.

3. How do you determine if a set is linearly independent in R^3?

To determine if a set is linearly independent in R^3, you can use the method of Gaussian elimination to reduce the vectors to row-echelon form. If there are no free variables and the rank of the matrix is equal to the number of vectors, then the set is linearly independent.

4. What is the importance of linear independence?

Linear independence is important because it allows us to determine the number of independent variables in a system of equations, which can help in solving the system. It also helps in understanding the geometric properties of vectors and their transformations.

5. Can a set be linearly independent in one vector space and linearly dependent in another?

Yes, a set can be linearly independent in one vector space and linearly dependent in another. This is because linear independence is dependent on the specific vector space and its dimensionality. A set may be linearly independent in R^3 but linearly dependent in R^2.

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