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Math Help - Coefficient Matrix Question

  1. #1
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    Coefficient Matrix Question

    Here is the question I am having difficulty with:

    If the echelon form of the coefficient matrix of a homogeneous system of linear equations is:

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

    Find a set of basic solutions of this system

    Any tips on where to start?
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  2. #2
    Member Ruun's Avatar
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    You have 3 equations and 5 unknows, so the dimension of the solution has 5-3 = 2 degrees of freedom (This is true for every system of linear equations. Number of unkowns minus number of linearly indepedent equations = degrees of freedom or dimension of the solution.) Think for example x_1=\lambda and x_2=\mu and solve it as a function of  \lambda and \mu
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  3. #3
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    Quote Originally Posted by dr4g0n9 View Post
    Here is the question I am having difficulty with:

    If the echelon form of the coefficient matrix of a homogeneous system of linear equations is:

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

    Find a set of basic solutions of this system

    Any tips on where to start?
    Just set up the equations corresponding to that matrix: I am going to call the unknown numbers x, y, z, u, and v. The bottom non-zero row gives the equation u+ 2v= 1 so u= 1- 2v. The row just above that is z+ u+ 2v= 0 so z= -u- 2v and since u= 1- 2v, z= -(1- 2v)- 2v= -1+ 2v- 2v= -1. Finally, the top row gives x+ 2y+ 3z+ 2v= 1. Since z= -1, that becomes x+ 2y- 3+ 2v= 1 or x= 4-2y- 2v. Thus, we have all unknown numbers written in terms of y and v.

    Another way to do this is to continue the row-reduction getting "0"s above the diagonal. Subtract three times the second row from the first row to get \begin{bmatrix} 1 & 2 & 0 & -3 & -4 & 1 \\0 & 0 & 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0\end{bmatrix}

    Now subtract the third row from the second to get \begin{bmatrix} 1 & 2 & 0 & -3 & -4 & 1 \\0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0\end{bmatrix}.

    Finally, subtract 3 times the third row from the first row to get 1 & 2 0 & 0 & 2 & 4 \\ 0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{bmatrix}

    Now the rows correspond to equations x+ 2y+ 2v= 4 so x= 4- 2y- 2v, z= -1, and u+ 2v= 1 so u= 1- 2v as before.
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  4. #4
    Super Member Gamma's Avatar
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    You can also just sort of look by inspection, you want to look from bottom up. You are looking for a 6x1 vector that satisfies this matrix equation being 0.

    The bottom two rows give you no information. Row three you need to find a linear combination so that a1 + b2 + c1=0. (-1,1,-1) seems like a good choice, so make those your last 3 entries in your vector.

    Now take that and look at the next row and see what you need to do to make it continue to work.
    1a + 1(-1) + 2(1) =0 \Rightarrow a=-1

    So now your vector looks like <*,*, -1, -1, 1, -1> and you just have 1 equation left to satisfy.

    1a + 2b + 3(-1) + 0(-1) + 2(1) + 1(-1) = 0 \Rightarrow a + 2b -2 =0 \Rightarrow a + 2b = 2

    <-2,2, -1, -1, 1, -1> works.

    Oh I just thought you needed a solution, didn't realize you needed the whole set of solutions, should have read more carefully.
    Last edited by Gamma; June 3rd 2009 at 11:57 AM. Reason: Whoops, misread question, but might be of use, so I left it.
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