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 anonymity Aug31-11 03:21 PM

Linear Algebra (applied) question

"Find all solutions to the augmented system"

1 2 0 3 1 ||-2
0 0 1 2 4 || 5
0 0 0 0 0 || 0
0 0 0 0 0 || 0

We just learned transposes and matrix multiplication (after a few sections on solving linear systems of equations -- via triangular matrices, row echelon form, and reduced row echelon form).

I applied the consistency theorem for linear systems, and was indeed able to write it as a linear combination of the column vectors of A -- so i know it's solvable.

The answer is given in the back of the book, and is given as the transpose of a row vector b (where Ax = b, and the above augmented matrix is (A | b)

Can anyone nudge me in the right direction? My book has no similar examples or mention of solving a system like this via transposes, and it seems that the transpose plays a roll in how I am SUPPOSED to solve this question...

ps: the answer is b = (8,-7,-1,7)T

maybe they just used this notation to save space...and I'm just supposed to use the method we have already learned for matrices in reduced row echelon form?

 vela Aug31-11 03:28 PM

Re: Linear Algebra (applied) question

It's just notation to save space.

 anonymity Aug31-11 03:38 PM

Re: Linear Algebra (applied) question

Okay, so that clears that up. However, looking at the problem with that in mind, there is no finite solution via the reduced row echelon method (the last two rows are all zero, so there will be two arbitrary constants in the solution). This does not match up with the solution given, which is made up of all real numbers...

So I would still maintain that I am missing something...

Would you agree?

 vela Aug31-11 03:41 PM

Re: Linear Algebra (applied) question

Can you post the problem as originally stated? Your post has some inconsistencies, like what exactly the vector b equals.

 Ray Vickson Aug31-11 03:41 PM

Re: Linear Algebra (applied) question

Quote:
 Quote by anonymity (Post 3478805) "Find all solutions to the augmented system" 1 2 0 3 1 ||-2 0 0 1 2 4 || 5 0 0 0 0 0 || 0 0 0 0 0 0 || 0 We just learned transposes and matrix multiplication (after a few sections on solving linear systems of equations -- via triangular matrices, row echelon form, and reduced row echelon form). I applied the consistency theorem for linear systems, and was indeed able to write it as a linear combination of the column vectors of A -- so i know it's solvable. The answer is given in the back of the book, and is given as the transpose of a row vector b (where Ax = b, and the above augmented matrix is (A | b) Can anyone nudge me in the right direction? My book has no similar examples or mention of solving a system like this via transposes, and it seems that the transpose plays a roll in how I am SUPPOSED to solve this question... ps: the answer is b = (8,-7,-1,7)T maybe they just used this notation to save space...and I'm just supposed to use the method we have already learned for matrices in reduced row echelon form?
The augmented system just reads as:
x1 + 2x2 + 3x4 + x5 = -2
x3 + 2x4 + 4x5 = 5
sum 0*xi = 0
sum 0*xi = 0.
You can solve for some of the variables in terms of the others.

RGV

 anonymity Aug31-11 03:52 PM

Re: Linear Algebra (applied) question

Quote:
 Quote by vela (Post 3478842) Can you post the problem as originally stated? Your post has some inconsistencies, like what exactly the vector b equals.
That is exactly how it is written. The matrix I have written at the top of my post is (A|b), and the solution is given as b = ()T

"Let Ax = b be a linear system whose augmented matrix (A|b) has reduced row echelon form [BLAH].

a) Find all solutions to the system
"

To Ray: that is sort-of the dilema. They say "find all solutions to the system" and give a non-ambiguous answer made up of only real constants, whereas what you propose will leave two arbitrary constants in the solution (which are not in the solution given by the book).

 vela Aug31-11 03:56 PM

Re: Linear Algebra (applied) question

Are you sure you're looking at the answer to the correct problem? :wink: Normally, you solve for x, not b, and clearly, b in the augmented matrix doesn't equal b in the answer, so something doesn't make sense here.

By the way, you have a 5-dimensional domain and two independent equations, so you'll have three arbitrary constants in the solution.

 anonymity Aug31-11 03:58 PM

Re: Linear Algebra (applied) question

Quote:
 Quote by vela (Post 3478862) By the way, you have a 5-dimensional domain and two independent equations, so you'll have three arbitrary constants in the solution.
Yeah I just realized that and felt dumb =|

and yes, it is absolutely the same question. I just checked while I was waiting for a response...

 anonymity Aug31-11 03:59 PM

Re: Linear Algebra (applied) question

This CANT be solved for x without introducing the arbitrary constants though...right?

 vela Aug31-11 04:17 PM

Re: Linear Algebra (applied) question

Right.

 anonymity Aug31-11 04:58 PM

Re: Linear Algebra (applied) question

Okay. There was actually a second part to the question on the next page, so the solution in the back must be for that part of the question (though it wasnt labled "b)").

Thanks for helping.

-Anonymous

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