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Matrix Help Please !

  1. Aug 7, 2005 #1
    Ok, i missed the class on finding the inverse of a matrix, and i only have a little bit of an idea on exactly what row operations i can do, when i try to make the matrix = its identity.

    Another question im stuck on.
    Q.

    I have 2 , 3 X 3 matrixs B and C respectivly.
    The question is find A if AB=C, and i know B and C
    Now, i know i cannot divide matrix's, and im stuck as to what way i should travel to find the matrix A.
    Thanks guys ! :shy:
     
  2. jcsd
  3. Aug 7, 2005 #2

    VietDao29

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    Here's what you should do. You have:
    [tex](AB)C = A(BC)[/tex]
    so:
    [tex]AB = C \Leftrightarrow (AB)B ^ {-1} = CB ^ {-1} = A(BB ^ {-1}) = A[/tex]
    So:
    [tex]A = CB ^ {-1}[/tex]
    So all you need is to find the inverse of B, and do a multiplication [tex]CB ^ {-1}[/tex]
    Viet Dao,
     
  4. Aug 7, 2005 #3
    Thanks, can you explain the first bit :surprised
    Did you use a rule to do that ?? or just decide to multiply one side by C, and one by BA :surprised
     
  5. Aug 7, 2005 #4

    TD

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    "Tools" you use:
    - associativity: [itex]A\left( {BC} \right) = ABC = \left( {AB} \right)C[/itex]
    - inverse (& identity-matrix): [itex]AA^{ - 1} = I = A^{ - 1} A[/itex]
    - property of the identity-matrix: [itex]AI = IA = A[/itex]

    [tex]\begin{gathered}
    AB = C \hfill \\
    {\text{multiply both sides at the right by }}B^{ - 1} : \hfill \\
    ABB^{ - 1} = CB^{ - 1} \Leftrightarrow A\left( {BB^{ - 1} } \right) = CB^{ - 1} \hfill \\
    {\text{since }}BB^{ - 1} = I: \hfill \\
    AI = CB^{ - 1} \hfill \\
    {\text{since }}AI = A: \hfill \\
    A = CB^{ - 1} \hfill \\
    \end{gathered}[/tex]
     
  6. Aug 7, 2005 #5
    TD, I have only a two-hour class on matrix.
    Multiplication of matrix can multiply on the right ?
     
  7. Aug 7, 2005 #6

    TD

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    Yes, you can multiply either right or left. In an equation though, you have to multiply both sides at the same place, so either both left or both right. So:
    [tex]\begin{gathered}
    A = B \Leftrightarrow AC = BC \to {\text{OK!}} \hfill \\
    A = B \Leftrightarrow CA = CB \to {\text{OK!}} \hfill \\
    A = B \Leftrightarrow AC = CB \to {\text{NOT OK!}} \hfill \\
    A = B \Leftrightarrow CA = BC \to {\text{NOT OK!}} \hfill \\
    \end{gathered} [/tex]
     
  8. Aug 7, 2005 #7
    For a matrix [tex]\left( \begin{array}{ccc}a&b\\c&d\end{array}\right)[/tex] the inverse of it will be [tex]\frac{1}{ad - bc} \left( \begin{array}{ccc}d&-b\\-c&a\end{array}\right)[/tex]

    This means that if matrix [tex]A[/tex] is equal to [tex]\left( \begin{array}{ccc}a&b\\c&d\end{array}\right)[/tex] then [tex]A A^{-1}[/tex] is equal to the identity matrix (as TD said)

    e.g. [tex]A A^{-1} = \left( \begin{array}{ccc}a&b\\c&d\end{array}\right) \times \frac{1}{ad - bc} \left( \begin{array}{ccc}d&-b\\-c&a\end{array}\right) = I[/tex]

    [tex]I = \left( \begin{array}{ccc}1&0\\0&1\end{array}\right)[/tex]

    The Bob (2004 ©)
     
  9. Aug 7, 2005 #8
    Does that work for 2X2 Matrix's only ??
     
  10. Aug 7, 2005 #9

    TD

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    Indeed, but in addition to Bob's explanation, this may be extended to larger matrices as followed:

    [tex]A^{ - 1} = \frac{1}
    {{\det \left( A \right)}}\operatorname{adj} \left( A \right)[/tex]

    Note that you divide by [itex]{\det \left( A \right)}[/itex] so it is clear that for the inverse to exist, [itex]\det \left( A \right) \ne 0[/itex] must be true. Also, [itex]\operatorname{adj} \left( A \right)[/itex] stands for the Adjugate Matrix.

    Another way is to extend the matrix to the right with the identity matrix and then swap (Gaussian Elimination), to get the identity left and the inverse on the right.
     
    Last edited: Aug 7, 2005
  11. Aug 7, 2005 #10
    Identity matrix must be a Square matrix
     
  12. Aug 7, 2005 #11
    Is there a proof for A times I = A ??
    I like to see a proof so the theory is concreted in my mind :eek:
     
  13. Aug 7, 2005 #12

    TD

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    As far as I know, this follows from definition! [itex]IA \equiv A[/itex]

    It's not that hard to see, if you remember how I looks like:

    [tex]\left( {\begin{array}{*{20}c}
    1 & 0 & \cdots & 0 \\
    0 & 1 & \cdots & 0 \\
    \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & \cdots & 1 \\

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

    It's a bit like asking a proof for why [itex]1 \cdot a = a[/itex], [itex]a \ne 0[/tex] of course.
     
  14. Aug 7, 2005 #13
    Yer, i was doing the second method, but im not sure what i do to the idenity matrix, and what rules and how when i multiply by a integer how it effects the identity etc. Like i know i need to make matrix A = to the identity, and do whatver i do to the matrix a, i have to do to the identity. But im not sure what i really do to it.
     
  15. Aug 7, 2005 #14
    AHH i just remember that I = 1 correct ?? If so that makes sense. Wait maybe im thinking of the determinant of I to be equal to 1 :uhh: :rofl:
    Infact, is it because when you end up multiyplying out with the identity, since both upper and lower regions are = to 0, you multiply something lol im so lost :shy: Atleast i try :surprised :yuck:
     
  16. Aug 7, 2005 #15

    TD

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    Well, you could see "I as 1", but then the "Matrix 1". All zero but 1's on the diagonal. [itex]\det \left( I \right) = 1[/itex] is correct too, not difficult to see!
     
  17. Aug 7, 2005 #16

    TD

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    Well, if you start with a 2x2 matrix, you 'add' to the right the 2x2 identity. I usually separate them with a dashed line or something, but when doing operation on it you should just see it as a larger 2x4 matrix as a whole. By applying the operation, you try to get the identity in the first 2x2 part, where A was. When you have that, you got the inverse of A on the right side, where the identity was.

    I wanted to post an example here for you but LaTeX can't generate it.
    Here's a link :smile:
     
    Last edited: Aug 7, 2005
  18. Aug 7, 2005 #17
    Thanks everyone, i completed the assignment :cool:
    I should be posting a few more little questions here :approve: Thanks :tongue:
     
  19. Aug 7, 2005 #18

    TD

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    Good to hear!
     
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