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Thread: 2x2 matrix diagonalisation formula.

  1. #1
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    2x2 matrix diagonalisation formula.

    Hi again overlords.

    Looking for some clarification regarding this question. I know how to diagonalise a standard 2x2 matrix just following the process, however I'm having a problem with understanding the formula as provided by the solution to this question. Clearly, to obtain their formula they're just multiplying matrices. The thing confusing me a little is the 2^n+1 step, which I have highlighted. I can see they are just multiplying the two left most matrices, but I don't understand why this input is not 4^n. Furthermore, how is the input provided (2^n+1) even equivalent to 4^n? if you sub in some value for n greater than 1, then 4^n > 2^n+1.

    2x2 matrix diagonalisation formula.-q3.png2x2 matrix diagonalisation formula.-3a.png

    Thanks in advance.
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  2. #2
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    Re: 2x2 matrix diagonalisation formula.

    Hey nalo6451.

    Hint - Try multiplying the first two matrices together and see if you get the same result as the next step.

    Show your work here if you want a more detailed response.
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  3. #3
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    Re: 2x2 matrix diagonalisation formula.

    2(2^n)= 2^{n+1} not 4^n
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  4. #4
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    Re: 2x2 matrix diagonalisation formula.

    Oh MY BAD guys, I forgot that I was supposed to add the exponents as they have the same base (pray for me).
    Last edited by nalo6451; Jun 14th 2017 at 01:37 AM.
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  5. #5
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    Re: 2x2 matrix diagonalisation formula.

    I didn't want to ask again but I actually have a related problem on the following question. I've highlighted the exact multiplication step I'm referring to. Now, I know 1/2 is equivalent to 2^-1, so I'm trying to multiply 4^n with 2^n-1. I can't figure out why they've collected the powers (n-1) with the 4.

    Thanks again friends.
    Attached Thumbnails Attached Thumbnails 2x2 matrix diagonalisation formula.-help-me-god.png  
    Last edited by nalo6451; Jun 14th 2017 at 02:59 AM.
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  6. #6
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    Re: 2x2 matrix diagonalisation formula.

    $\dfrac{1}{2}=\dfrac{2}{4}=2\cdot 4^{-1}$

    $\dfrac{1}{2}\cdot 4^n = 2\cdot 4^{-1}\cdot 4^n=2\cdot 4^{n-1}$
    Last edited by SlipEternal; Jun 14th 2017 at 03:47 AM.
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