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Math Help - 3x4 Matrix

  1. #1
    Member ~berserk's Avatar
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    3x4 Matrix

    Can someone help me solve this systems of equations problem using a matrix and in reduced row echelon form:

    3x-y+z=10
    2x-4y-z=20
    6x+8y+z=-100

    i got the answers to be x: -6, y=-8, and z=0

    i need help,though, on how to complete this problem without a calculator, like a walk through. any help would be greatly appreciated

    i have it set up here
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  2. #2
    Member ~berserk's Avatar
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    anyone?
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  3. #3
    Member maybeline9216's Avatar
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    "i got the answers to be x: -6, y=-8, and z=0'

    Are these the correct answers??
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  4. #4
    Member maybeline9216's Avatar
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    "i got the answers to be x: -6, y=-8, and z=0"

    These answers are totally wrong!!!
    To check the answers, just sub their respective values inside each of the equations...Here's the workings i have written manually...Please click on the attached file and click on it again and then zoom in to have a better view.But my workings are rather long though and besides, u can just use mental calculation^_^ without using calculator
    Attached Thumbnails Attached Thumbnails 3x4 Matrix-matrix.jpg  
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  5. #5
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    typo in line 2 of matrix led to error - should be -1
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  6. #6
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    Hi,

    With a system of equations you can add multiples of one equation to another. This way you can find a solution much quicker than you have already.

    For example, if you multiply your first equation by 2 you get 6x-2y+2z=20.

    if you subtract this equation from the 3rd equation you get 6y+z=120.

    By continuing this process you will eventually reduce an equation with just one unknown. You should be able to do this after 2 or 3 steps.

    The other method (which in principle is identical to this approach) is to represent the system of equations into a matrix and do what is called "gaussian elimination" which is a smart approach to take - have a look on the net for a description.

    Hope this helps!
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  7. #7
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    to get an upper triangular form of the matrix

    1) keep row one
    subtract/add multiples of this row to achieve zeros in the rest of column one

    2) keep row 2
    subtract/add multiples of this row to achieve zeros in the rest of column two

    1) in this case R2-2/3R1
    R3-2R1

    2) R3+3R2


    GIVES 3 -1 1 : 10
    0 -10/3 -5/3 :40/3
    0 0 -6 : 40

    -6Z = 40
    Z=20/3 SO Y = -22/3 SO X = -4/3

    Hope my arithmetic is right!
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  8. #8
    Member ~berserk's Avatar
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    I know that i can use Gaussian elimination but I'm supposed to solve it with a 3x4 matrix or else i would have had it done a loooooooong time ago

    this is the goal for the problem

    1 0 0 : x
    0 1 0 : y
    0 0 1 : z


    also sorry, I do see the mistake I made it is supposed to be:
    3 -1 1 : -10
    2 -4 -1 : 20
    6 8 1 :-100
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  9. #9
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    Thumbs up Use matrix online calculator!

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  10. #10
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    Hello, ~berserk!

    \begin{array}{ccc} 3x-y+z &=& 10 \\ 2x-4y-z &=& 20 \\ 6x+8y+z& =& \text{-}100 \end{array}

    We have: . \left|\begin{array}{ccc|c}3 & \text{-}1 & 1 & 10 \\ 2 & \text{-}4 & \text{-}1 & 20 \\ 6 & 8 & 1 & \text{-}100 \end{array}\right|


    \begin{array}{c}R_1-R_2 \\ \\ \\ \end{array} \left|\begin{array}{ccc|c} 1 & 3 & 2 & \text{-}10 \\ 2 & \text{-}4 & \text{-}1 & 20 \\ 6 & 8 & 1 & \text{-}100 \end{array}\right|


    \begin{array}{c}\\ R_2-2R_1 \\ R_3-6R_1 \end{array} \left|\begin{array}{ccc|c} 1 & 3 & 2 & \text{-}10 \\ 0 & \text{-}10 & \text{-}5 & 40 \\ 0 & \text{-}10 & \text{-}11 & \text{-}40 \end{array} \right|


    \begin{array}{c} \\ \text{-}\frac{1}{10}R_2 \\ R_3-R_2 \end{array} \left|\begin{array}{ccc|c} 1 & 3 & 2 & \text{-}10 \\ 0 & 1 & \frac{1}{2} & \text{-}4 \\ 0 & 0 & \text{-}6 & \text{-}80 \end{array} \right|


    \begin{array}{c} R_1-3R_2 \\ \\ \text{-}\frac{1}{6}R_3 \end{array} \left| \begin{array}{ccc|c}1 & 0 & \frac{1}{2} & 2 \\ 0 & 1 & \frac{1}{2} & \text{-}4 \\ 0 & 0 & 1 & \frac{40}{3} \end{array} \right|


    \begin{array}{c} R_1-\frac{1}{2}R_3 \\ \\[-4mm] R_2 - \frac{1}{2}R_3 \\ \\ \end{array} \left| \begin{array}{ccc|c}1 & 0 & 0 & \text{-}\frac{14}{3} \\ \\[-4mm] 0&1&0 & \text{-}\frac{32}{3} \\ \\[-4mm] 0&0&1 & \frac{40}{3} \end{array} \right|


    Therefore: . \begin{Bmatrix} x&=&\text{-}\dfrac{14}{3} \\ \\[-3mm] y &=& \text{-}\dfrac{32}{3} \\ \\[-3mm] z &=& \dfrac{40}{3} \end{Bmatrix}

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