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Inverse matrices help please

  1. Apr 26, 2006 #1
    hi there would anyone be able to give me a link or guidence as how i would go about finding the inverse of A? I have a book that tells me add subtract and multiply.

    the question is this: find the inverse of A

    A =

    (19 81)
    (2 10)

    sorry for the crude matrices. any help would be great :)

    lakitu
     
  2. jcsd
  3. Apr 26, 2006 #2
    there is a trick for finding the inverse of a 2x2 matrix
    [tex]A=\left( \begin{array}{cc} a & b \\
    c & d \end{array} \right)[/tex]
    then
    [tex]A^{-1}=\frac{1}{ad-bc}\left( \begin{array}{cc} d & -b \\
    -c & a \end{array} \right)[/tex]
    provided ad-bc is not equal to zero
    hope that helps
     
  4. Apr 26, 2006 #3
  5. Apr 27, 2006 #4
    thank you for your comments :)
     
  6. Apr 27, 2006 #5

    HallsofIvy

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    If your book is telling you "add, subtract, and multiply" (hey, that's how you solve any mathematics problem!:rolleyes: ) then go back and read over exactly what you add and subtract and what you multiply by. I suspect that your book is talking about "row operations"- that's what Gale's website is talking about.
     
    Last edited: Apr 28, 2006
  7. Apr 27, 2006 #6

    robphy

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    Before rushing off to a formula or procedure, it might be helpful to understand what you are trying to do when determining the inverse of a matrix.

    Given a square matrix [tex]A=\left( \begin{array}{cc} a & b \\
    c & d \end{array} \right)[/tex],
    its inverse [if it exists] is another square matrix [tex]A^{-1}=\left( \begin{array}{cc} p & q \\
    r & s \end{array} \right)[/tex] such that the matrix product is the identity matrix.

    [tex]
    \begin{align*}
    AA^{-1}&=I\\
    \left( \begin{array}{cc} a & b \\
    c & d \end{array} \right)
    \left( \begin{array}{cc} p & q \\
    r & s \end{array} \right)
    &=
    \left( \begin{array}{cc} 1 & 0 \\
    0 & 1\end{array} \right)
    \end{align*}
    [/tex]
    If you carry out the matrix multiplication, you should find a simple system of four linear equations in four unknowns... "simple" because it's really a pair of systems of two linear equations in two unknowns. You can easily solve these systems to obtain the formula given by vladimir69 above.
    (In addition, the inverse would satisfy [tex]A^{-1}A&=I[/tex] as well.)
     
  8. Apr 28, 2006 #7

    HallsofIvy

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    Of course, if you have a 3 by 3 or 5 by 5 matrix, so that your system is 9 equation in 9 unknowns or 25 equations in 25 unknowns, you might find robphy's method a bit tedious! I think it's worth learning row reduction.
     
  9. Apr 28, 2006 #8
    i think it's a bit more trickier than just plain trick, it's actually a theorem.
     
  10. Apr 28, 2006 #9

    robphy

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    Agreed!... assuming one first understands what it means to take the inverse of a matrix.
     
  11. Apr 28, 2006 #10

    Hurkyl

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    This is how I like to think of it. If this doesn't make sense to you, feel free to forget about it so you don't get confused!


    We start off with the partitioned matrix:

    [A : I]

    which has the property that: (the left matrix) = (the right matrix) * A. In particular, A = I*A.

    Now, if we do row operations, we will get some other partitioned matrix:

    [B : C]

    which still has the property that: (the left matrix) = (the right matrix) * A. In particular, B = C*A

    If we fully row-reduce the left hand side, we get the partitioned matrix:

    [I : V]

    which still has the property that (the left matrix) = (the right matrix) * A. In particular, I = V*A, and therefore V is the inverse of A.
     
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