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A matrix multiplied by it's inverse is the identity matrix, right?

  1. Jan 28, 2010 #1
    Matrix A= 2x2, R1= -1, -1, R2= -7, 3
    Matrix b= 2x2, R1= 1,0, R2= 0, 1

    A*?=b
    ____________

    To solve, I put ? on the one side of the equation as ?=A^(-1)b. My answer is then just the inverse of A, because what is multiplied by the identity matrix is itself. It is shown to be incorrect.

    ?=2x2, R1= -3, -1, R2= -7, -1


    Please help! I would like this to become more intuitive.
    Also, is there a way to format matrices on the forum?
     
  2. jcsd
  3. Jan 28, 2010 #2

    vela

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    How did you calculate the inverse? The matrix you wrote for ? isn't the inverse of A.
     
  4. Jan 28, 2010 #3

    LCKurtz

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    You just forgot to divide by the determinant of A when you calculated the inverse. You can click on the equation below to see how to write a matrix in tex.

    [tex]A^{-1}=\left [
    \begin{array}{cc}
    -\frac 3 4&-\frac 1 4\\

    -\frac 7 4&-\frac 1 4

    \end{array}
    \right ][/tex]
     
  5. Jan 29, 2010 #4
    Thank you for your help.

    I did forget to divide the matrix by its determinate, 1/[ad-bc]. It would then be my previous answer divided by -1/10, right?

    It turned up incorrect, still. I also tried your answer, but still the same response.
     
  6. Jan 29, 2010 #5

    LCKurtz

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    The determinant is -10, but you don't "divide by -1/10". You divide by -10. So you should get

    [tex]A^{-1} = -\frac {1} {10}\left[ \begin{array}{cc}
    3&1\\
    7&-1

    \end{array}
    \right ][/tex]
     
  7. Jan 29, 2010 #6
    That is what I meant. The answer is still wrong.
     
  8. Jan 29, 2010 #7

    LCKurtz

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    [tex]
    A^{-1} = -\frac {1} {10}\left[ \begin{array}{cc}
    3&1\\
    7&-1

    \end{array}
    \right ]
    [/tex]

    is the correct inverse. If you are trying to solve a system like

    Ax = b

    where x and b are column vectors then the solution will be

    x = A-1b
     
  9. Jan 29, 2010 #8

    vela

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    Are you sure you're using the same matrix as LCKurtz? Your original inverse not only was missing the determinant factor but had the wrong sign on one of the elements as well.
     
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