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Quick way of raising matrices to indicies

by thomas49th
Tags: indicies, matrices, raising
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thomas49th
#1
Nov30-09, 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
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Ben Niehoff
#2
Nov30-09, 03:21 PM
Sci Advisor
P: 1,594
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.
thomas49th
#3
Nov30-09, 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 :(

Billy Bob
#4
Nov30-09, 04:12 PM
P: 392
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.
thomas49th
#5
Nov30-09, 05:00 PM
P: 656
aha so the formula for that specific elemnt is [tex]\sin 2^{n-1}x[/tex]

Correct?

Thanks
Tom
Billy Bob
#6
Nov30-09, 07:34 PM
P: 392
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.)
thomas49th
#7
Dec3-09, 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 :(
Billy Bob
#8
Dec3-09, 03:07 PM
P: 392
What is the trig identity for cos(A+B)? For sin(A+B)?

Also, your A^3 has two incorrect entries.
thomas49th
#9
Dec3-09, 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
Borek
#10
Dec3-09, 04:17 PM
Admin
Borek's Avatar
P: 23,602
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.
Billy Bob
#11
Dec3-09, 05:08 PM
P: 392
Quote Quote by thomas49th View Post

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

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

Additional hint: A(x) corresponds to rotation through an angle of x.
thomas49th
#12
Dec3-09, 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!!!
Billy Bob
#13
Dec3-09, 09:18 PM
P: 392
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!
HallsofIvy
#14
Dec4-09, 04:58 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,569
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.
thomas49th
#15
Dec6-09, 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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