Quick way of raising matrices to indicies

  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. Ben Niehoff

    Ben Niehoff 1,755
    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. 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. Since 2 cos x sin x = sin 2x, a pattern is emerging, just not the one you thought.
  6. aha so the formula for that specific elemnt is [tex]\sin 2^{n-1}x[/tex]


  7. 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. 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. What is the trig identity for cos(A+B)? For sin(A+B)?

    Also, your A^3 has two incorrect entries.
  10. 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. Borek

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

    Additional hint: A(x) corresponds to rotation through an angle of x.
  13. I can see A^2 = sin2x

    but I cannot see why that is so argghh! must be something really simple!!!
  14. A=
    [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. HallsofIvy

    HallsofIvy 41,055
    Staff Emeritus
    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. 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!
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?