# Quick way of raising matrices to indicies

by thomas49th
Tags: indicies, matrices, raising
 P: 648 1. The problem statement, all variables and given/known data A given matrix $$A(x) =\left | \begin{array}{cc} cos(x)&sin(x)\\ -sin(x)&cos(x) \end{array}\right |$$ 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? Thanks Thomas
 Sci Advisor P: 1,516 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.
 P: 648 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 :(
P: 394

## 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.
 P: 648 aha so the formula for that specific elemnt is $$\sin 2^{n-1}x$$ Correct? Thanks Tom
 P: 394 Maybe for $$A^{2^{n-1}}$$ but that doesn't help with $$A^5$$. In other words, multiply $$A^2$$ by $$A$$, not by $$A^2$$, to find $$A^3$$. (I know you know this -- your brain just went into overdrive and caused a blunder.)
 P: 648 ahhh Im getting A^3 $$A(x) =\left | \begin{array}{cc} cos(x)cos(2x)-sin(x)sin(2x)& cos(x)cos(2x) +sin(x)sin(2x)\\ -cos(x)cos(2x)-sin(x)sin(2x)& cos(x)cos(2x)-sin(x)sin(2x) \end{array}\right |$$ where is this pattern? I can see they're all similar I might as well just expand to A^5 now :(
 P: 394 What is the trig identity for cos(A+B)? For sin(A+B)? Also, your A^3 has two incorrect entries.
 P: 648 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
 Admin P: 21,703 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.
P: 394
 Quote by thomas49th 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.
 P: 648 I can see A^2 = sin2x but I cannot see why that is so argghh! must be something really simple!!!
 P: 394 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!
 PF Patron Sci Advisor Thanks Emeritus P: 38,412 You might also note that $$\begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & sin(\theta)\end{bmatrix}$$ 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.
 P: 648 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

 Related Discussions General Physics 3 Linear & Abstract Algebra 7 Advanced Physics Homework 3 General Math 7 Introductory Physics Homework 1