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Math Help - Row-Reduced Echelon Form / Vector-Parametric Solution

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    Row-Reduced Echelon Form / Vector-Parametric Solution

    We are asked to find the Row-Reduced Echelon Form (RREF) of some augmented coefficient matrices for systems of linear equations. I know how to apply the elementary row operations but what I don't know is that given a question, how you should approach it and what are the correct row operations to perform. Is there a strategy to approach with or is it just trial and error?

    Here is the matrix:

    \left(\begin{array}{ccccccl}5 & 1 & 7 & -1 & 2 & 6 & |23\\ 2 & 9 & -23 & -2 & -18 & 25 & |-1\\ 1 & 3 & -7 & 2 & -3 & 14 & |4\\ 6 & 2 & 6 & 1 & 3 & 14 & |29\end{array}\right)

    Apologies for the augmented section, I couldn't find the right tex for it. Hope it's clear. Anyway I began to approach it by interchanging rows 3 and 1 to get the leading 1 in the top left, then subtracting multiples of the new row 1 to get zeros below, but it got to a point where there wasn't a clear next step. As I understand the RREF is unique, correct? So I don't think I'll neccessarily be able to find it by trial and error?

    Also, given the RREF - I am asked to find the vector-parametric solution but again this is not in my notes, nor can I find it online.
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    Re: Row-Reduced Echelon Form / Vector-Parametric Solution

    yes, you may perform the elementary row-operations in any order you wish. if you don't mind some messy fractions, this is what i do:

    1. multiply the first row so that the first entry is 1 (if the first entry is 0, swap with a row that has a non-zero first entry first).

    2. make the first column 0, below the 1.

    3. make the second entry in the second row 1 (if it is 0, swap with some other row below it that has a a non-zero 2nd entry).

    4. clear out the second column, except for the leading one in row 2 (it may happen that the 2nd column is already all 0, in which case i turn my attention to columns 3).

    5. repeat, moving diagonally down.

    you may see some short-cuts along the way that cause you to work "out-of-order" (such as two rows being equal or obvious multiples at an early stage).

    your (unaugmented) matrix has 6 columns and 4 rows, which translates to 6 variables and 4 equations. this system is "under-determined" so it is likely you will have to arbitrarily set one or more variables to something, in order to get values for the rest. these variables are "free variables", as you can choose them how you please- once chosen, the other variables are determined, so they are "bound variables". it is common to denote the free variables as parameters and write s,t,u,.... etc. for them, given the other bound variables in terms of these. for example,

    x1 + x3 = 3
    x2 - x3 = 0

    (this corresponds to the rref:

    \begin{bmatrix}1&0&1&3\\0&1&-1&0\\0&0&0&0 \end{bmatrix}, )

    is "free" in x3, and any solution can be written (3-t,t,t) or (3,0,0) + t(-1,1,1).
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    Re: Row-Reduced Echelon Form / Vector-Parametric Solution

    Thanks a lot.

    Seeing as we already have a row with a one in the first column, is it not better to interchange this with the top row to start off with and then proceed with making zeros? It reduces the amount of calculations to make by 1.

    Also, what determines whether there are zero rows or not, and whether you'll be able to get a one in the 'next' column? (diagonally down from the previous leading 1?). Since there's more columns not every one will have a leading 1 so I'm not sure how to determine which will. Would some calculations cause you to need to double back on yourself and not be able to lead to a rref?

    Also, for the solution: how do you determine which variables to use as the free variables?
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    Re: Row-Reduced Echelon Form / Vector-Parametric Solution

    if you keep moving "right and down" you'll always make progress.

    after multiplying the first row by suitable numbers, and adding to the other rows to clear out the rest of the first column, no subsequent row-operation (except swapping rows) will change the first column. so when we get that leading 1, with all 0's below it, that "locks" the first row into place, and no matter how we swap or modify the remaining rows, the first column stays the same. whether the leading 1 in the second row turns out to be in the second, third, or later position, it will be rightward of the first 1 in the first row. in other words, we keep making progress on getting all 0's below the diagonal.

    although it is possible to choose the "free" variables in many different ways, it seems most efficient to me to work "from the bottom up". so if your last variable (that isn't always 0) is xj, i set that to "t", let's say. hopefully the "next variable up" is determined by an equation like:

    axi + bxj = c,

    so you have xi = (c - bxj)/a.

    but the next equation up, when you get to the rref form, may have 3 terms, in which case you need a second parameter.

    when you get the rref form for your matrix, post what you have and i'll give a better explanation.
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    Re: Row-Reduced Echelon Form / Vector-Parametric Solution

    Thanks again, you're a great help.

    What I meant about knowing when there is a leading one is, what will tell you whether a column will contain a leading one or not? Say you manage to get one in the first two columns (which I have), but not in the third column, how will you know for sure the third column doesn't have one? I have a zero in the third row, third column (diagonally down from two leading ones). Does this tell me to move on to the next column straight away and continue to proceed through with the method?

    Also, would you advise when making leading ones to add multiples of other rows or just multiply by 1 over whatever's in that place (what you meant by messy fractions I'm assuming). I got to a stage with -639, 190, 33 and 3424 in a column and I'm trying to turn the 3424 into a 1. Should I just divide by 3424? (although if I used this method I guess I wouldn't be in this position). It's easier than solving -639x + 190y = 3423.
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    Re: Row-Reduced Echelon Form / Vector-Parametric Solution

    The solution is:

    x_{1} + 2x_{3} + x_{5} + x_{6} = 5
    x_{2} - 3x_{3} - 2x_{5} + 3x_{6} = -1
    x_{4} + x_{5} + 2x_{6} = 1

    One zero row, rank 3 with 6 variables. So we have 3 free variables? How would I write this in vector parametric form please?

    I know the first step is write in terms of the pivot variables:

    x_{1} = 5 - 2x_{3} - x_{5} - x_{6}
    x_{2} = 3x_{3} +2x_{5} - 3x_{6} -1
    x_{4} = 1 - x_{5} - 2x_{6}

    Then I'm a little unsure of what to do next.
    Last edited by Shizaru; October 16th 2011 at 10:59 AM.
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    Re: Row-Reduced Echelon Form / Vector-Parametric Solution

    assign parameters to x3,x5 and x6 (s,t and u are the traditional choices). then express x1, x2, and x4 in terms of s,t,and u....it will look like this:

    (5-2s-t-u,3s+2t-3u-1,s,1-t-u,t,u) although it looks cleaner if you write it like so:

    s(-2,3,1,0,0,0) + t(-1,2,0,-1,1,0) + u(-1,-3,0,-1,0,1) + (5,-1,0,1,0,0,0) (these are often written as column vectors)

    your solution set is a 3-dimensional subspace "translated" by a constant vector =

    general solution of homogeneous system + one particular solution of non-homogeneous system.

    note that (5,-1,0,1,0,0,0) is indeed "a" solution: 5(5) +(-1)(1) + 0(7) + 1(-1) + 0(2) + 0(6) = 23

    5(2) + (-1)(9) + 0(-23) + 1(-2) + 0(-18) + 0(25) = -1,

    5(1) + (-1)(3) + 0(-7) + 1(2) + 0(-3) + 0(14) = 4

    5(6) + (-1)(2) + 0(6) + 1(1) + 0(3) + 0(14) = 29
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    Re: Row-Reduced Echelon Form / Vector-Parametric Solution

    Quote Originally Posted by Deveno View Post
    assign parameters to x3,x5 and x6 (s,t and u are the traditional choices). then express x1, x2, and x4 in terms of s,t,and u....it will look like this:

    (5-2s-t-u,3s+2t-3u-1,s,1-t-u,t,u) although it looks cleaner if you write it like so:

    s(-2,3,1,0,0,0) + t(-1,2,0,-1,1,0) + u(-1,-3,0,-1,0,1) + (5,-1,0,1,0,0,0) (these are often written as column vectors)

    your solution set is a 3-dimensional subspace "translated" by a constant vector =

    general solution of homogeneous system + one particular solution of non-homogeneous system.

    note that (5,-1,0,1,0,0,0) is indeed "a" solution: 5(5) +(-1)(1) + 0(7) + 1(-1) + 0(2) + 0(6) = 23

    5(2) + (-1)(9) + 0(-23) + 1(-2) + 0(-18) + 0(25) = -1,

    5(1) + (-1)(3) + 0(-7) + 1(2) + 0(-3) + 0(14) = 4

    5(6) + (-1)(2) + 0(6) + 1(1) + 0(3) + 0(14) = 29
    Thanks very much once again,

    I just want to ask, is there a difference in this case between the existing variables and the assigned parameters, i.e. insteading of using s,t and u, can we not just continue using x3, x5, x6? I recall in the examples we covered, the solution was in the form as you gave but only using the initial variables to express it. Nothing like s,t, u etc, just x(number)s.

    It looked more like this:

    "(5,-1,0,1,0,0) + x3(-2,3,1,0,0,0) + x5(-1,2,0,-1,1,0) + x6(-1,-3,0,-1,0,1)"

    Does this still make sense, as I believe this is the format we are expected to use. (I'm not questioning what you've written, just asking if they are comparable, hope this makes sense!)
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    Re: Row-Reduced Echelon Form / Vector-Parametric Solution

    that's perfectly OK. the symbols used for the parameters can be anything. doing it like you have there emphasizes which variables (from your original set) are the "non-pivots".
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