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Homework Help: Quick way of raising matrices to indicies

  1. Nov 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A given matrix [tex]
    A(x) =\left | \begin{array}{cc} cos(x)&sin(x)\\ -sin(x)&cos(x) \end{array}\right |

    Find the term in the first row and second column A^{5}(x) when x = 18°

    2. Relevant equations

    3. The attempt at a solution

    Instead of multiplying the matrix again and again by itself in a iterative procedure is there any quicker way?

  2. jcsd
  3. Nov 30, 2009 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    For this particular matrix, why don't you find A^2 for general x, and use your knowledge of trigonometric identities...then you may see a pattern.
  4. Nov 30, 2009 #3
    A^2 = [ cos2x 2cosxsinx
    -2cosxsinx cos2x ]

    what I want is cos²x + sin²x so i get one, but what I got at the mooment doesn't seem to get me any where :( I wanted somthing like if the power is odd then you get -1 and even 1 in the first element, but can't see that happening :(
  5. Nov 30, 2009 #4
    Since 2 cos x sin x = sin 2x, a pattern is emerging, just not the one you thought.
  6. Nov 30, 2009 #5
    aha so the formula for that specific elemnt is [tex]\sin 2^{n-1}x[/tex]


  7. Nov 30, 2009 #6
    Maybe for [tex]A^{2^{n-1}}[/tex] but that doesn't help with [tex]A^5[/tex].

    In other words, multiply [tex]A^2[/tex] by [tex]A[/tex], not by [tex]A^2[/tex], to find [tex]A^3[/tex].

    (I know you know this -- your brain just went into overdrive and caused a blunder.)
  8. Dec 3, 2009 #7
    ahhh Im getting

    A(x) =\left | \begin{array}{cc} cos(x)cos(2x)-sin(x)sin(2x)& cos(x)cos(2x) +sin(x)sin(2x)\\ -cos(x)cos(2x)-sin(x)sin(2x)& cos(x)cos(2x)-sin(x)sin(2x) \end{array}\right |


    where is this pattern? I can see they're all similar
    I might as well just expand to A^5 now :(
  9. Dec 3, 2009 #8
    What is the trig identity for cos(A+B)? For sin(A+B)?

    Also, your A^3 has two incorrect entries.
  10. Dec 3, 2009 #9
    cos(A+B) = cosACosB - sinAsinB
    sin(A+B) = sinAcosB + cosAsinB


    I can see how we can use RCos(x+a) for A^3 but I want to get A^5. There must be some relationship!

    Which of the terms are incorrect in A^3

    btw for A^2 i got [cos2x sin2x
    -sin2x cos2x]

  11. Dec 3, 2009 #10


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    Staff: Mentor

    I am sure BB means something much faster, but whenever I hear about anything raised to large integer power I think in terms of powers of 2, in this case it means A5=A*(A2)2 - that means just three multiplications instead of 5. The larger the exponent, the more you save.
  12. Dec 3, 2009 #11

    Can't you guess the correct pattern just from this! :smile:

    Additional hint: A(x) corresponds to rotation through an angle of x.
  13. Dec 3, 2009 #12
    I can see A^2 = sin2x

    but I cannot see why that is so argghh! must be something really simple!!!
  14. Dec 3, 2009 #13
    [cos x sin x
    -sin x cos x]

    [cos 2x sin 2x
    -sin 2x cos 2x]

    what would you guess! Yes, it's very simple, sorry. :smile:

    Now compute A^3, and if it doesn't match your guess, then your computation is probably wrong, not your guess!
  15. Dec 4, 2009 #14


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    Science Advisor

    You might also note that
    [tex]\begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & sin(\theta)\end{bmatrix}[/tex]
    corresponds to a rotation by -x degrees. Multiplying it by itself 5 times is the same as rotating by -5x degrees. In this case that is -5(18)= -90 degrees. The result is suprisingly simple.
  16. Dec 6, 2009 #15
    yeah I didn't think of that!

    still need to expand it manually to see the maths but bed time now

    Thanks for the help!
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