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Solving a system of equations by Gauss's method

  1. Feb 15, 2016 #1
    1. The problem statement, all variables and given/known data
    So I have the following :

    AX=B, where

    $$A=\begin{bmatrix}
    a_{ij}
    \end{bmatrix}_{mXn}$$

    with $$a_{ij}=i$$

    $$
    X=\begin{bmatrix}
    X_{1} &X_{2} &X_{3} &... &X_{n}
    \end{bmatrix}^T$$

    and

    $$B=\begin{bmatrix}
    1 &2 &3 &... &m
    \end{bmatrix}^T$$

    2. Relevant equations

    Gauss' method of solving system of linear equations

    3. The attempt at a solution

    Now the thing is, when I usually solve a system of equation with Gauss I usually get 3 normal equations. It's the first time I get this and I'm not even sure of what I have in my hands...

    Could somebody help?
     
  2. jcsd
  3. Feb 16, 2016 #2

    Svein

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    If [itex]a_{i,j}=i [/itex] then the two first rows are [itex]
    \begin{pmatrix}
    1 & 1 & ... \\
    2 & 2 & ... \\
    ... & ... & ... \\
    \end{pmatrix}
    [/itex] What does that tell you about the determinant of A?
     
  4. Feb 16, 2016 #3

    SteamKing

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    It's not clear what you mean when you say you get 3 normal equations when you use Gauss' method. Normal equations are formed when you multiply A by its transpose, and this method of forming normal equations is frequently used to solve for the unknown coefficients in a regression equation. Even then, the system of normal equations must be solved using some technique, but because the resulting system is usually only a 2 x 2 or a 3 x 3, Cramer's Rule can be employed rather than Gaussian elimination.

    Your present system of equations is a little more general. The matrix of coefficients A is composed of elements where each element in a particular row is equal to its row number. That's what ##a_{ij} = i## means. Ditto for the RHS: the value of each element in the B vector is also equal to the row number.

    Thus, your system looks like this (for n = 3):

    Code (Text):


    | 1   1   1 |   | X1 |    | 1 |
    | 2   2   2 | * | X2 |  = | 2 |
    | 3   3   3 |   | X3 |    | 3 |

     
    Applying Gaussian elimination to this system of equations will produce an upper triangular matrix of coefficients with just a single non-zero coefficient in the n-th equation. The n-th unknown can then be solved and the values of the other n-1 unknowns can be determined in turn by proceeding from the n-th equation back to the first equation, using back-substitution.

    The technique is illustrated here:

    http://mathworld.wolfram.com/GaussianElimination.html
     
  5. Feb 16, 2016 #4
    So I'm supposed to solve this generally ? I just meant that it's the first time I get something like this. Never saw this.

    How do I even begin this in the first place ? I know how to use Gauss, it's just the setting up that's causig problems.
     
  6. Feb 16, 2016 #5
    I drew this : http://imgur.com/5lQZNvk

    How do I get my system now ? I m going to go sleep, I hope I get an answer :(
     
  7. Feb 16, 2016 #6

    SteamKing

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    Look at the clue given to you by Svein in Post #2.

    What can you say about the determinant of A in general, given its composition?
     
  8. Feb 16, 2016 #7
    Well it's going to be the "i" that we place in it. like 1,1,1,1...,1

    2,2,2,...,2

    etc.

    Oh I didn't notice his post. Sorry
     
  9. Feb 16, 2016 #8
    I didn't lean about determinants yet
     
  10. Feb 16, 2016 #9

    SteamKing

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    That makes it difficult to determine if a system of linear equations has a solution, if you don't know anything about determinants.
     
  11. Feb 16, 2016 #10
    So there's no other way for me to determine the answer ???
     
  12. Feb 16, 2016 #11

    SteamKing

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    I don't know what to tell you.

    Studying matrix algebra without studying determinants is like studying arithmetic but omitting discussions of multiplication.

    If you don't know what a determinant is, Cramer's Rule is off the table, for example, since that method is based entirely on calculating determinants.

    While you don't need to explicitly calculate the determinant of the matrix of coefficients to use Gaussian elimination, knowledge that the determinant is not zero suggests that there is a unique solution to the associated system of linear equations.
     
  13. Feb 16, 2016 #12
    I didn't say we won'T learn it, we just didn't at the moment. Now what I wonder is why did I get this problem if it's not possible for me to solve it without any tools...
     
  14. Feb 16, 2016 #13

    SteamKing

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    Well, skip determinants for the time being.

    Hve you learned how to tell if a system of linear equations is independent?
     
  15. Feb 16, 2016 #14

    Ray Vickson

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    You do not need determinants; you just need to actually know how Gaussian elimination works. Then you need to carry out the actual Gaussian elimination steps. I mean: do it instead of agonizing about it.

    To help you get going, I suggest very strongly that you do it from start to finish first for the case of n = 2 and then for the case of n = 3. Both of these are small enough that you should have no difficulties. Come back with more questions (if you need to) after you have done what I suggested.
     
  16. Feb 16, 2016 #15
    Ok I'll go try and see if i can figure it out.
     
  17. Feb 16, 2016 #16
    ok, so for 2x2 I get the following equations :

    x1+x2=1

    2x1+2x2=2

    These are good right ? I multiplied the matrices
     
  18. Feb 16, 2016 #17

    SteamKing

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    To use elimination, you don't need to multiply the matrices together. It's better if you leave them as original.

    Maybe you missed reading this link:

    http://mathworld.wolfram.com/GaussianElimination.html

    But it contains an example of how to use elimination to solve a system of equations. You should study this example carefully.
     
  19. Feb 16, 2016 #18
    Ok thanks, going to read this. It looks like what I have.
     
  20. Feb 17, 2016 #19

    Svein

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    Then divide all elements in row 2 by 2. What do you get?
     
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