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Linear Equation Methods

  1. Aug 5, 2010 #1
    For the system of linear algebraic equations:
    x − y = 1
    2x + 3y = 7
    a) find solution by the Cramer’s rule;
    b) find solution by the method of Gaussian elimination;
    c) write the first iteration of the Jacobi method with the initial guess x^0=y^0=1.

    I understand Cramer's Rule, and I think I understand Gaussian elimination, have got
    x=2 and y=1 for both of them, but I have no idea how to do part c). Can anyone help me?
     
  2. jcsd
  3. Aug 5, 2010 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    In terms of matrices, this system would be written as
    [tex]\begin{bmatrix}1 & -1 \\ 2 & 3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}1 \\ 7\end{bmatrix}[/tex]

    We can take out the "diagonal" and rewrite it as
    [tex]\begin{bmatrix}1 & 0 \\ 0 & 3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}i+ \begin{bmatrix}0 & -1 \\2 & 0\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}1 \\ 7\end{bmatrix}[/tex]

    so that
    [tex]\begin{bmatrix}1 & 0 \\ 0 & 3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}1 \\ 7\end{bmatrix}- \begin{bmatrix}0 & -1 \\2 & 0\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}[/tex]

    or, just in terms of the equations, x= 1+ y, 3y= 7- 2x so we have x= 1+ y, y= (7- 2x)/3.

    Taking [itex]x^0= y^0= 1[/itex], we have [itex]x^1= 1+ y^0= 1+ 1= 2[/itex] and [itex]y^1= (7- 2x^0)/3= (7- 2)/3= 5/3[/itex].
     
  4. Aug 5, 2010 #3
    Thank you, very helpful! :biggrin:
     
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