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Math Help - gaussian elimination method

  1. #1
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    gaussian elimination method

    Am I doing this correctly for matices?
    x has subscripts 1 and 2

    -2x
    + 2x = 2
    3x - 3x = -3

    |-2, 2, 2 |
    |3, -3, -3|

    -2(-1\2) = 1 so

    |1, -1, -1|
    |3, -3, -3|

    -3(1) + 3 = 0 so

    |1, -1, -1|
    |0, 0, 0 |

    The unique solution is x subcript 1 = ? and x subscript 2 = ?
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  2. #2
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    You only need to find the inverse of the coefficient matrix

    \displaystyle \left[<br />
  \begin{array}{ c c }<br />
     -2 & 2 \\<br />
     3 & -3<br />
  \end{array} \right]<br />

    But does the inverse for this matrix exist?
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  3. #3
    Senior Member Sambit's Avatar
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    Quote Originally Posted by hansfordmc View Post

    |1, -1, -1|
    |0, 0, 0 |

    The unique solution is x subcript 1 = ? and x subscript 2 = ?
    [/SIZE]
    Whenever you get a only series of zeros in the final step of Gaussian Elimination, it implies  0.x_1+0.x_2=0 which is obvious and holds for any values of x_1,x_2. In such case you do not get any unique solution, what you get is a family of solutions. In your problem, the family of solution is given by \{(x_1,x_2)|x_2-x_1=1\}
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  4. #4
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    Hello, hansfordmc!

    Am I doing this correctly for matices?

    . . \begin{array}{ccc}\text{-}2x + 2y &=& 2 \\ 3x - 3x &=& \text{-}3 \end{array}


    \text}We have: }\;\left| \begin{array}{cc|c}\text{-}2 & 2 & 2 \\<br />
3 & \text{-}3 & \text{-}3 \end{array}\right|


    . . . . \begin{array}{c}\text{-}\frac{1}{2}R_1 \\ \\ \end{array}\left|\begin{array}{cc|c} 1 & \text{-}1 & \text{-}1 \\ 3 & \text{-}3 & \text{-}3 \end{array}\right|


    \begin{array}{c} \\ R_2-3R_1 \\ \end{array}\left| \begin{array}{cc|c} 1 & \text{-}1 & \text{-}1 \\ 0 & 0 & 0 \end{array}\right| . This is correct!


    \text{The unique solution is: }\:\begin{Bmatrix}x\;=\;? \\ y \;=\;?\end{Bmatrix}

    As Sambit pointed out, a row of zeros indicates no unique solution.

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  5. #5
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    So now, exactly what was the question? What you have shown so far is that both equations, -2x_1+ 2x_2= 2 and 3x_1- 3x_2= -3 reduce to x_1- x_2= -1 or x_1= x_2- 1. If the question was "is there a unique solution", the answer is "no". If the problem was to find all possible solutions, then (x_1, x_2)= (t- 1, t), is a solution for any number, t, and all solutions are of that form.
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