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Homework Help: Finding x and y in matrix

  1. May 12, 2008 #1
    Can someone tell me what i'm doing wrong. the answer for x is supposed to be -3/7 and y is supposed to be 5/7 but i have 5/7 and -23/7.

    [tex]


    \begin{bmatrix}
    1 & -2 \\
    3 & 1
    \end{bmatrix}
    \begin{bmatrix}
    1 & 0 \\
    0 & 1
    \end{bmatrix}



    [/tex]
    new r1= r1-(r2 x -2)
    [tex]


    \begin{bmatrix}
    7 & 0 \\
    3 & 1
    \end{bmatrix}
    \begin{bmatrix}
    1 & -2 \\
    0 & 1
    \end{bmatrix}



    [/tex]
    new r1=r1/7
    [tex]


    \begin{bmatrix}
    1 & 0 \\
    3 & 1
    \end{bmatrix}
    \begin{bmatrix}
    1/7 & -2/7 \\
    0 & 1
    \end{bmatrix}



    [/tex]
    new r2 = r2-(r1 x 3)
    [tex]


    \begin{bmatrix}
    1 & 0 \\
    0 & 1
    \end{bmatrix}
    \begin{bmatrix}
    1/7 & -2/7 \\
    -3/7 & 13/7
    \end{bmatrix}



    [/tex]

    [tex]

    b=\begin{bmatrix}
    1 \\
    -2
    \end{bmatrix}
    [/tex]
    so
    [tex]

    \begin{bmatrix}
    1/7 & -2/7 \\
    -3/7 & 13/7
    \end{bmatrix}
    \begin{bmatrix}
    1 \\
    -2
    \end{bmatrix}



    [/tex]

    [tex]
    4/7 + 1/7 = 5/7
    [/tex]
    [tex]
    -26/7 + 3/7 = -23/7
    [/tex]
     
    Last edited: May 12, 2008
  2. jcsd
  3. May 12, 2008 #2

    Defennder

    User Avatar
    Homework Helper

    Are you sure y is supposed to be 5/7? I got -5/7. As for your error, check your first step. The matrix on the left hand side should be 1,2 not 1,-2.
     
  4. May 12, 2008 #3
    yes its negative 5/7 actually.
     
  5. May 12, 2008 #4

    HallsofIvy

    User Avatar
    Science Advisor

    It would help if you would tell us right at the beginning what the problem is!
    I can guess, as Defennnder apparently did, that you are trying to solve
    [tex]\begin{bmatrix}1 & -2 \\3 & 1\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}= \begin{bmatrix}1 \\ -2\end{bmatrix}[/tex]
    and that you found the inverse matrix then mutliplied the inverse by b.

    Here's your error. In the last column "r1- (r2*-2)" would be 0- (1(-2))= 2, not -2. it would have been simpler to think "r1= r1+ (r2*2)" rather than having the two negatives.




     
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