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Math Help - matrices help!

  1. #1
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    matrices help!

    I'm teaching myself matrices because it could prove useful in the future.

    I'll just run through a few things I've done to check if I'm correct:

    i'm confident in adding/subtracting and multiplying.

    What I'm having trouble getting my mind around is identities, inverses and determinants for 2x2 matrices.

    In my book it says that for matrices of a given dimension (2x2) a identity I exists...

    \left(\begin{array}{cc}1&0\\0&1\end{array}\right)

    What does this I mean?

    Then after this it says

    If A =\left(\begin{array}{cc}2&3\\1&4\end{array}\right) , AI = IA = A

    I have no idea in the world what this means. Could someone clarify it for me please =]

    Thanks in advanced!
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  2. #2
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    The matrix I has the same function to matrices as the number 1 has to numbers.

    If you multiply any number by one and you will get the same number back.

    If you multiply any matrix by I and you will get the same matrix back.

    Nothing more to it!
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  3. #3
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    Oh okay! Cool stuff.

    Now... there's an example question here which I don't really get...

    For the Matrix A = \left(\begin{array}{cc}3&2\\1&6\end{array}\right) find: A^-1

    For the solution they wrote:
     A^-1 = \frac{1}{16} \left(\begin{array}{cc}3&2\\1&6\end{array}\right)

    how does that work?
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  4. #4
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    Quote Originally Posted by jgv115 View Post
    Oh okay! Cool stuff.

    Now... there's an example question here which I don't really get...

    For the Matrix A = \left(\begin{array}{cc}3&2\\1&6\end{array}\right) find: A^-1

    For the solution they wrote:
     A^{-1} = \frac{1}{16} \left(\begin{array}{cc}3&2\\1&6\end{array}\right)

    how does that work?
    (Put the -1 in { } to get the whole thing as the exponent.)
    If you multiply [tex]AA^{-1}= \left(\begin{array}{cc}3&2\\1&6\end{array}\right) \left(\begin{array}{cc}3&2\\1&6\end{array}\right)= \left(\begin{array}{cc}16 & 0 \\ 0 & 16\end{array}\right)[/quote]
    The "1/16" in front gives the indentity matrix- the determinant of A is 16 and that is one over the determinant. In other words, "I" is like "1" in the real numbers and " A^{-1} is like 1/A.

    There are a variety of different ways to find the inverse of a matrix (provided it has one- many matrices do not have inverses) and I don't know which your text gives.

    My preference is: Write A and the identity matrix side by side. Use a series of "row operations" to reduce A to the indentity matrix while simultaneously applying those same row operations to the indentity matrix. When A has been reduced to the indentity matrix, the identity matrix will have been change to A^{-1}.
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  5. #5
    MHF Contributor Calculus26's Avatar
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    See attachment
    Attached Thumbnails Attached Thumbnails matrices help!-invrse.jpg  
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  6. #6
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    Quote Originally Posted by jgv115 View Post
    Oh okay! Cool stuff.

    Now... there's an example question here which I don't really get...

    For the Matrix A = \left(\begin{array}{cc}3&2\\1&6\end{array}\right) find: A^-1

    For the solution they wrote:
     A^-1 = \frac{1}{16} \left(\begin{array}{cc}3&2\\1&6\end{array}\right)

    how does that work?
    Use the formula and work out on your own instead, the solution provided is wrong.
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