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Math Help - Kindly Help me plss!!! Matrices

  1. #1
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    Kindly Help me plss!!! Matrices

    1 0 0 K1
    0 1 0 k2
    0 0 1 k3

    3.) x - 3y -7z = 6
    4x + y = 7
    2x + 3y +z = 9

    4.) 3x - 5z = -1
    2x + 7y = 6
    x + y + z = 5

    ... Help T_T this is trial & error probs ... so many, so complicated ... kindly teach me some tips in solving matrices?
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  2. #2
    dan
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    if the matrix is a 3 x 3 it there is a simple way to solve...
    a b c
    D= d e f = aei + dhc +gbf - ceg-fha- idb. at least i think that's it^_^
    g h i

    please note this in only for a 3X3 matrix

    dan
    Last edited by dan; September 9th 2006 at 04:39 PM.
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  3. #3
    dan
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    for a larger matrix you have to cofacter...do a forum serch for "quicklopedia"
    i think he talked about that there...

    dan
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  4. #4
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    Quote Originally Posted by dan View Post
    for a larger matrix you have to cofacter...do a forum serch for "quicklopedia"
    i think he talked about that there...

    dan
    No, for a larger matrix you use elementary row operations to row reduce it to echelon form. The complexity of this algorithm is n^2, the complexity of using the determinant becomes larger not exactly sure.
    What I am trying to say, is that for larger matrices using the determinant is not the most efficient way to do it.

    To example, to find the inverse matrix using Gaussian elimination takes n^3 steps. To find the inverse using transposes and determinants takes n! ways thus for the first values we have,
    1,8,27,64,125...
    1,2,6,24,120,...
    You can see it is more efficient using Gaussian elimination for matrices of dimension 5.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    No, for a larger matrix you use elementary row operations to row reduce it to echelon form. The complexity of this algorithm is n^2, the complexity of using the determinant becomes larger not exactly sure.
    What I am trying to say, is that for larger matrices using the determinant is not the most efficient way to do it.

    To example, to find the inverse matrix using Gaussian elimination takes n^3 steps. To find the inverse using transposes and determinants takes n! ways thus for the first values we have,
    1,8,27,64,125...
    1,2,6,24,120,...
    You can see it is more efficient using Gaussian elimination for matrices of dimension 5.
    The problem looks to me like an exercise in Gaussian elimination.

    RonL
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  6. #6
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    ive got an example in our discussion:

    2x - y + 3z = 9
    3x + y + +2z = 11
    x - y + z = 2

    A
    ( Add 2nd row to the 1st row)
    2 -1 3 9
    3 1 2 11
    1 -1 1 2

    B
    (5 0 5 20) 1/5
    3 1 2 11
    1 -1 1 2

    C
    (Add row 2 by row 3)
    1 0 1 4
    3 1 2 11
    1 -1 1 2

    D
    (Add 2nd row by 1st row)
    (1 0 1 4) -2
    3 1 2 11
    4 0 3 13

    E
    ( Add 3rd row by 1st row)
    (1 0 1 4) -3
    -1 1 0 3
    4 0 3 13

    F
    (Add 3rd row by 1st row)
    1 0 1 4
    1 1 0 3
    (1 0 0 1) -1

    G
    (Add 2nd row by 3rd row)
    0 0 1 3
    1 1 0 3
    (1 0 0 1)-1

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

    I
    1 0 0 1 where x = 1
    0 1 0 2 where y = 2
    0 0 1 3 where z = 3

    looks simple but complicated
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  7. #7
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    Nevermind Ive already finished answering it
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  8. #8
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    ive got an example in our discussion:

    2x - y + 3z = 9
    3x + y +2z = 11
    x - y + z = 2
    Let me do it my way.
    As I said, all you need is to reach a triangluar matrix. (A matrix with the bottom numbers zeros below the diagnol).
    The augmented matrix is
    [2 -1 3 | 9]
    [3 1 2 | 11]
    [1 -1 1 | 2]
    Thus,
    [1 -1 1 | 2]
    [2 -1 3 | 9]
    [3 1 2 | 11]
    Thus,
    [1 -1 1 |2]
    [0 1 1 |5 ]
    [0 4 -1| 5 ]
    Thus,
    [1 -1 1|2]
    [0 1 1|5 ]
    [0 0 -5| -15]
    Thus,
    [1 -1 1|2]
    [0 1 1|5]
    [0 0 1|3]
    And stop right here, why? You will see, first note you got a triangular matrix.
    The bottom line says,
    z=3
    The middle line says
    y+z=5 thus y+3=5 thus y=2
    The top line says,
    x-y+z=2 thus x-2+3=2 thus x+1=2 thus x=1.
    Finished!
    You see how each line leads to a simple equation.
    Doing it like this would say half your time.

    This is my 24th post!!!
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    Let me do it my way.
    As I said, all you need is to reach a triangluar matrix. (A matrix with the bottom numbers zeros below the diagnol).
    The augmented matrix is
    [2 -1 3 | 9]
    [3 1 2 | 11]
    [1 -1 1 | 2]
    Thus,
    [1 -1 1 | 2]
    [2 -1 3 | 9]
    [3 1 2 | 11]
    Thus,
    [1 -1 1 |2]
    [0 1 1 |5 ]
    [0 4 -1| 5 ]
    Thus,
    [1 -1 1|2]
    [0 1 1|5 ]
    [0 0 -5| -15]
    Thus,
    [1 -1 1|2]
    [0 1 1|5]
    [0 0 1|3]
    And stop right here, why? You will see, first note you got a triangular matrix.
    The bottom line says,
    z=3
    The middle line says
    y+z=5 thus y+3=5 thus y=2
    The top line says,
    x-y+z=2 thus x-2+3=2 thus x+1=2 thus x=1.
    Finished!
    You see how each line leads to a simple equation.
    Doing it like this would say half your time.

    This is my 24th post!!!
    Stopping when the matrix is upper-triangular and then using back substitution
    is fine if we just have the one equation to solve. But if we have a number of
    right-hand sides (as we get if we are using Gaussian elimination to numerically
    invert a matrix) it may be more efficient to go all the way and reduce the left-
    hand side to an identity matrix.

    It may be that this exercises is intended to illustrate the technique for
    subsequent application in this manner, so if the original poster has been
    asker to reduce the left-hand side to the identity matrix we should respect
    that.

    RonL
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  10. #10
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    Quote Originally Posted by CaptainBlack View Post
    Stopping when the matrix is upper-triangular and then using back substitution
    is fine if we just have the one equation to solve. But if we have a number of
    right-hand sides (as we get if we are using Gaussian elimination to numerically
    invert a matrix) it may be more efficient to go all the way and reduce the left-
    hand side to an identity matrix.

    It may be that this exercises is intended to illustrate the technique for
    subsequent application in this manner, so if the original poster has been
    asker to reduce the left-hand side to the identity matrix we should respect
    that.

    RonL
    That can only happen if that system is not "well-behaved".
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  11. #11
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    Quote Originally Posted by ThePerfectHacker View Post
    That can only happen if that system is not "well-behaved".
    Whatever the merits of the algorithm (and no algorithm is perfect) the OP
    seems to being asked to do it that way.

    RonL
    Last edited by CaptainBlack; September 10th 2006 at 10:58 AM.
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  12. #12
    dan
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    ph, i'de like to learn more about the 5x5 matrix solving with  n^3 and [tex]n![tex]
    WHAT IS RONG WITH THE CODE ON THE SITE?
    do you have a link, or what would i search for on google??
    dan
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  13. #13
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    Quote Originally Posted by dan View Post
    WHAT IS RONG WITH THE CODE ON THE SITE?
    LaTeX is down because of repairs.
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  14. #14
    dan
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    Quote Originally Posted by ThePerfectHacker View Post
    LaTeX is down because of repairs.
    ok thanks,


    by the way...i thought it was, "there are only 10 types of people: those who know binary and those who don't".
    mayby you use a different version in NY than we use in NE

    `dan
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  15. #15
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    Quote Originally Posted by dan View Post
    ok thanks,


    by the way...i thought it was, "there are only 10 types of people: those who know binary and those who don't".
    mayby you use a different version in NY than we use in NE
    Try to stay on topic please, during these past days there was not off topic talk. Let us keep it down.

    Just to answer you, that is the third time someone told me that joke is supposed to be ten, yes, I know. But the joke it that it is not.
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