
#1
Aug3111, 03:21 PM

P: 163

"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? 



#2
Aug3111, 03:28 PM

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PF Gold
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It's just notation to save space.




#3
Aug3111, 03:38 PM

P: 163

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? 



#4
Aug3111, 03:41 PM

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Linear Algebra (applied) question
Can you post the problem as originally stated? Your post has some inconsistencies, like what exactly the vector b equals.




#5
Aug3111, 03:41 PM

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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 



#6
Aug3111, 03:52 PM

P: 163

"Let Ax = b be a linear system whose augmented matrix (Ab) has reduced row echelon form [BLAH]. a) Find all solutions to the system " To Ray: that is sortof the dilema. They say "find all solutions to the system" and give a nonambiguous 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). 



#7
Aug3111, 03:56 PM

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Are you sure you're looking at the answer to the correct problem? 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 5dimensional domain and two independent equations, so you'll have three arbitrary constants in the solution. 



#8
Aug3111, 03:58 PM

P: 163

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



#9
Aug3111, 03:59 PM

P: 163

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




#10
Aug3111, 04:17 PM

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Right.




#11
Aug3111, 04:58 PM

P: 163

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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