help in easy algerbra queston...

[Semidevil]

this is actually a linear algebra question.

ok, so I am learning about linear dependence, and linear independence.

the example gives problem gives me the following matrix(lets call it matrix A):

1 2 0 0
1 5 1 0
-2 -1 1 0

the problem says that by gaus elimation, this reduces to(lets call this matrix B)

1 2 0 0
0 1 1/3 0
0 0 0 0

and this says that this matrix is linear dependent because there is a non trivial solution.

I"m having troulbe understanding why matrix A reduced to matrix B. Couldn't we reduce it some more, and at the end get a cleaner answer, like 1 non-zero number on each row??

 

[yoda291]

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[Heisenberg]

Some matrices can only be reduced so far. If you only had one non-zero value on each row then most likely the system would be linearly independent because only the trivial solution would exist.

 

[Ness]

what is the matrix?

 

[Semidevil]

Originally posted by: Heisenberg
Some matrices can only be reduced so far. If you only had one non-zero value on each row then most likely the system would be linearly independent because only the trivial solution would exist.

I think my problem is that I am having trouble understanding between linearly independent and dependent.

from my book and my understanding, if each row has only 1 or none non-zero number, then it is independent, and if it has more then 1 non zero number in its row, then it is dependent.

is that all?

I"m actually having troulbe reducing matrixes...is there a technique? I usually just do it column by column...is that correct?

 

[Heisenberg]

Originally posted by: Semidevil

Originally posted by: Heisenberg
Some matrices can only be reduced so far. If you only had one non-zero value on each row then most likely the system would be linearly independent because only the trivial solution would exist.

I think my problem is that I am having trouble understanding between linearly independent and dependent.

from my book and my understanding, if each row has only 1 or none non-zero number, then it is independent, and if it has more then 1 non zero number in its row, then it is dependent.

is that all?

I"m actually having troulbe reducing matrixes...is there a technique? I usually just do it column by column...is that correct?

I don't remember the actual procedure. I just use my calculator. Anyway, you basically have it right about the difference between dependent and independent. If only the trivial solution exists, then every variable will have to be zero to satisfy the equations and you'll only have one number in a row.

 

[Semidevil]

cool, thanx.....

I have a TI86...can my TI 86 reduce for me? haha

but anyways, I"m having trouble w/ when to stop reducing the matrix. like the example I put up. how does the book know to stop at the matrix

1 2 0 0
0 1 1/3 0
0 0 0 0

?

I think I can just keep going in circles and circles.......and just try to reduce the matrix......

 

[CrazyPerson]

if its easy. do it yourself

 

[Heisenberg]

Originally posted by: Semidevil
cool, thanx.....

I have a TI86...can my TI 86 reduce for me? haha

but anyways, I"m having trouble w/ when to stop reducing the matrix. like the example I put up. how does the book know to stop at the matrix

1 2 0 0
0 1 1/3 0
0 0 0 0

?

I think I can just keep going in circles and circles.......and just try to reduce the matrix......

Yeah, the 86 can do it. IIRC you stop when you have the matrix to where each row starts with a one and has only zeroes before that.

 

[Semidevil]

sorry to bother you, but can you tell me how to do it on the 86?

I tried "aug(B)" and I got an error(just random guess)

p.s: B is the name of the matrix...

so basically, I just stop whenever the ones are in the "correct place" in the rows?

 

[Heisenberg]

You can use either ref(B) for row echelon form or rref(B) for reduced row echelon form. rref reduces it a little further.
Yeah, you basically just stop when you have leading ones in each row.

 

[Semidevil]

ok thanx.

now, I swear, the very last question....

I need to write this vector as a linear comboination of vectors in the set

so the matrix is

1 0 -1 = 1
2 1 0 = 1
3 2 1 = 1

so I guess I augment this and reduce it to

1 0 -1 1
0 1 2 -1
0 0 0 0

the book says that from this, you can say that

x1 = 1 + t
x2 = -1 -2t
x3 = t

now, where did all the t's and numbers come from??

so lost here

 

[Heisenberg]

When you reduce the matrix you basically end up with two equations and three unknowns.
x1 - x3 = 1
x2 + 2*x3 = -1

(remember that the first three columns in the matrix are coefficients for your unknowns and the fourth column is the right-hand side of the equations)

So now you just pick a parameter (say t) and set one of the unknowns equal to it. So x3=t. Now you just solve for x1 and x2 in terms of x3 (which is t). What's happening is you have infinitely many solutions because you have three unknows and two equations so you just pick one variable to express the other two in.

 

[Crypticburn]

Originally posted by: Semidevil
ok thanx.

now, I swear, the very last question....

I need to write this vector as a linear comboination of vectors in the set

so the matrix is

1 0 -1 = 1
2 1 0 = 1
3 2 1 = 1

so I guess I augment this and reduce it to

1 0 -1 1
0 1 2 -1
0 0 0 0

the book says that from this, you can say that

x1 = 1 + t
x2 = -1 -2t
x3 = t

now, where did all the t's and numbers come from??

so lost here

x3 is arbitrary, x2 = -1-2*x3, and x1 = 1+x3. In that case, the book chose x3 to be equal to t (x3 = t)

Crypticburn

 

[Spencer278]

matrices just make my head spin why did some fool have to come up with such pain in the ass part of math.

 

[mAdD INDIAN]

Originally posted by: Spencer278
matrices just make my head spin why did some fool have to come up with such pain in the ass part of math.

lol. I find vector space and its related proofs quite difficult. It's because its too abstract for me. General matrices aren't bad.


Laplace transforms, now that's evil. As is polar coordinates...I frigging hate polar coordinates...especially with double integrals thrown in.