# Quick way of raising matrices to indicies

1. ### thomas49th

655
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

2. ### Ben Niehoff

1,663
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. ### thomas49th

655
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. ### Billy Bob

392
Since 2 cos x sin x = sin 2x, a pattern is emerging, just not the one you thought.

5. ### thomas49th

655
aha so the formula for that specific elemnt is $$\sin 2^{n-1}x$$

Correct?

Thanks
Tom

6. ### Billy Bob

392
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.)

7. ### thomas49th

655
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 :(

8. ### Billy Bob

392
What is the trig identity for cos(A+B)? For sin(A+B)?

Also, your A^3 has two incorrect entries.

9. ### thomas49th

655
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

### 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.

11. ### Billy Bob

392

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

Additional hint: A(x) corresponds to rotation through an angle of x.

12. ### thomas49th

655
I can see A^2 = sin2x

but I cannot see why that is so argghh! must be something really simple!!!

13. ### Billy Bob

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!

14. ### HallsofIvy

40,203
Staff Emeritus
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.

15. ### thomas49th

655
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