# Homework Help: Lin. Alg. - Is the set a linearly independent subset of R^3

1. Sep 4, 2007

### b0it0i

1. The problem statement, all variables and given/known data
Is {(1,4,-6), (1,5,8), (2,1,1), (0,1,0)} a linearly independent subset of R^3. Justify your answer

2. Relevant equations

3. 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

c = 14
d = 111
b = -13
a = -15

are 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

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.

Last edited: Sep 5, 2007
2. Sep 4, 2007

### proton

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. Sep 4, 2007

### Dick

a=-15, b=-13, c=14, d=111. Yes, you've missed some solutions, as proton predicted.

4. Sep 5, 2007

### b0it0i

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

c = 14
d = 111
b = -13
a = -15

are 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. Sep 5, 2007

### b0it0i

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. Sep 5, 2007

### Dick

You've got it. My solution was just a 'for instance'.