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Math Help - gaussian elimination

  1. #1
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    Question need help urgent !!

    Gaussin Elimation

    1)x+2y+6z=12
    2z=2
    -3x-6y-9z=-27

    2)x-y+z=22
    2y-z=-23
    2x+3y=-11.2

    3)x+2y+4z=7
    3x+y+4z=-2
    2x+9y-2x=10


    4)x+2y+4z=7
    3x+y+4z=-2
    2x+9y-2z=10
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  2. #2
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    Hello, cheesepie!

    #1 is a set of dependent equations.


    1)\;\begin{array}{ccc}x+2y+6z\:=\:12\\2z\:=\:2 \\ -3x-6y-9z\:=\:-27\end{array}

    We have: . \begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 2 & | & 2 \\ \text{-}3 & \text{-}6 & \text{-}9 & | & \text{-}27\end{vmatrix}

    \begin{array}{ccc} \\ R_2\div2\\ R_3\div(\text{-}3)\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 1 & 2 & 3 & | & 9\end{vmatrix}

    \begin{array}{ccc}\\ \\ R_3-R_1\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & \text{-}3 & | & \text{-}3\end{vmatrix}

    \begin{array}{ccc} \\ \\ R_3\div(-3)\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & 1 & | & 1\end{vmatrix}

    \begin{array}{ccc} \\ \\ R_3-R_2\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & 0 & | & 0\end{vmatrix}


    Since we have a rows of 0's (zeros all the way across), the system is dependent.

    There is an infinite number of solutions to the system.

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  3. #3
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    tks...how about other question ?
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  4. #4
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    Hello, cheesepie1

    I need to know exactly where your difficulty is . . .
    . . You don't understand Gaussian Elimination at all?
    . . You don't know how to get the 1's and 0's?
    . . You don't know what order to work at it?

    I can spend hours, doing ALL your homework for you,
    . . but there is no learning taking place, it's wasted effort.

    Having said that, here's the second one.
    And I would really like to know exactly why you can't do the third one.


    2)\;\begin{array}{ccc}x-y+z \:=\:22 \\ 2y - z\:=\:-23 \\ 2x + 3y\:=\:-11.2\end{array}

    We have: . \begin{vmatrix}1 & \text{-}1 & 1 & | & 22 \\ 0 & 2 & \text{-}1 & | & \text{-}23 \\ 2 & 3 & 0 & | & \text{-}11.2\end{vmatrix}


    \begin{array}{ccc} \\R_2\div2\\R_3-2\cdot R_1\end{array}\;\begin{vmatrix}1 & \text{-}1 & 1 & | & 22 \\ 0 & 1 & \text{-}0.5 & | & \text{-}11.5 \\ 0 & 5 & \text{-}2 & | & \text{-}55.2\end{vmatrix}


    \begin{array}{ccc}R_1+R_2\\ \\ R_3-5\cdotR_2\end{array}\;\begin{vmatrix}1 & 0 & 0.5 & | & 10.5 \\ 0 & 1 & \text{-}0.5 & | & \text{-}11.5 \\ 0 & 0 & 0.5 & | & 2.3\end{vmatrix}


    \begin{array}{cccc} R_1-0.5\cdot R_3 \\ R_2+0.5\cdot R_3 \\ \\ \end{array}\;\begin{vmatrix}1 & 0 & 0 & | & 8.2 \\ 0 & 1 & 0 & | & \text{-}9.2 \\ 0 & 0 & 1 & | & 4.6\end{vmatrix}


    Answers: . x = 8.2,\;y = -9.2,\;z = 4.6

    And remember, you can check your answers yourself.

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  5. #5
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    i dont understand the inverse matrices




    example how u change inverse of b =
    [ 8 4 ]
    [ 3 2 ]

    into

    [1 0 ]
    [0 1 ]
    Last edited by cheesepie; July 10th 2006 at 07:41 AM.
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  6. #6
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    You aren't inverting matrices using this method. You are performing "row operations." For example, you have the system of equations:
    (1) x - y + z = 22
    (2) 2y - z = -23
    (3) 2x + 3y = -11.2

    I will solve this in an identical manner as Soroban, except that he is using Gaussian elimination and I am simply manipulating equations. What I want you to note is that they are the same thing, just that one is in matrix notation and one is not.

    So. The first operation Soroban did was R2 \div2. This is the same as dividing both sides of equation 2 by 2:

    (1) x - y + z = 22
    (2)' y - 0.5z = -11.5
    (3) 2x + 3y = -11.2

    The next thing he did was R3 - 2R1. This is subtracting twice equation 1 by R3.
    2x + 3y = -11.2 (Equation 3)
    -2x + 2y - 2z = -44 (Equation 1 with both sides multiplied by -2 )
    ---------------------------
    (3)' 5y - 2z = -55.2

    So we have:
    (1) x - y + z = 22
    (2)' y - 0.5z = -11.5
    (3)' 5y - 2z = -55.2

    Now R1 + R2. This is equation 1 + equation 2'

    x - y + z = 22
    y -0.5z = -11.5
    -------------------
    (1)' x + 0.5z = 10.5

    So
    (1)' x + 0.5z = 10.5
    (2)' y - 0.5z = -11.5
    (3)' 5y - 2z = -55.2

    Now R3 - 5R2 means equation 3' - 5 times equation 2"
    5y - 2z = -55.2
    -5y +2.5z = 57.5
    ----------------------
    (3)" 0.5z = 2.3

    So we have

    (1)' x + 0.5z = 10.5
    (2)' y - 0.5z = -11.5
    (3)" 0.5z = 2.3

    R1 - R3 means equation 1' minus 0.5 times equation 3" (Soroban's statement here was incorrect.)
    x + 0.5z = 10.5
    - 0.5z = -2.3
    ----------------
    (1)" x = 8.2

    and R2 + R3 means equation 2' plus equation 3" (Soroban's statement here was incorrect.)
    y - 0.5z = -11.5
    0.5z = 2.3
    ----------------
    (2)"' y = -9.2

    So
    (1)" x = 8.2
    (2)"' y = -9.2
    (3)" 0.5z = 2.3 (Then, of course, solve this for z!)

    You can compare each step in the above with Soroban's matrix form.

    -Dan
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  7. #7
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    Hello, cheesepie!

    Now you've changed the subject . . .


    i dont understand the inverse matrices

    Example: how u change inverse of b =
    [ 8 4 ]
    [ 3 2 ]

    into

    [1 0 ]
    [0 1 ]

    You haven't been taught this procedure either??

    The set up is: / \begin{bmatrix}8 & 4 & | & 1 & 0 \\ 3 & 2 & | & 0 & 1\end{bmatrix}

    Now, use Gaussian Elimination to change the left matrix into the Identity: . \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}

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