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Thread: 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.


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

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

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

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

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

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


    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.


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

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


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


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


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


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

    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; Jul 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 $\displaystyle \div$2. 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: /$\displaystyle \begin{bmatrix}8 & 4 & | & 1 & 0 \\ 3 & 2 & | & 0 & 1\end{bmatrix}$

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

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