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Quick way of raising matrices to indicies 
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#1
Nov3009, 03:17 PM

P: 656

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  [/tex] 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? Thanks Thomas 


#2
Nov3009, 03:21 PM

Sci Advisor
P: 1,588

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.



#3
Nov3009, 04:00 PM

P: 656

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 :( 


#4
Nov3009, 04:12 PM

P: 393

Quick way of raising matrices to indicies
Since 2 cos x sin x = sin 2x, a pattern is emerging, just not the one you thought.



#5
Nov3009, 05:00 PM

P: 656

aha so the formula for that specific elemnt is [tex]\sin 2^{n1}x[/tex]
Correct? Thanks Tom 


#6
Nov3009, 07:34 PM

P: 393

Maybe for [tex]A^{2^{n1}}[/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.) 


#7
Dec309, 02:52 PM

P: 656

ahhh Im getting
A^3 [tex] 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  [/tex] where is this pattern? I can see they're all similar I might as well just expand to A^5 now :( 


#8
Dec309, 03:07 PM

P: 393

What is the trig identity for cos(A+B)? For sin(A+B)?
Also, your A^3 has two incorrect entries. 


#9
Dec309, 03:59 PM

P: 656

cos(A+B) = cosACosB  sinAsinB
sin(A+B) = sinAcosB + cosAsinB right? 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] Thanks Thomas 


#10
Dec309, 04:17 PM

Admin
P: 23,402

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 A^{5}=A*(A^{2})^{2}  that means just three multiplications instead of 5. The larger the exponent, the more you save.



#11
Dec309, 05:08 PM

P: 393

Can't you guess the correct pattern just from this! Additional hint: A(x) corresponds to rotation through an angle of x. 


#12
Dec309, 05:19 PM

P: 656

I can see A^2 = sin2x
but I cannot see why that is so argghh! must be something really simple!!! 


#13
Dec309, 09:18 PM

P: 393

A=
[cos x sin x sin x cos x] A^2= [cos 2x sin 2x sin 2x cos 2x] A^3= what would you guess! Yes, it's very simple, sorry. Now compute A^3, and if it doesn't match your guess, then your computation is probably wrong, not your guess! 


#14
Dec409, 04:58 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,361

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. 


#15
Dec609, 06:11 PM

P: 656

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! Thomas 


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