# Homework Help: Matrix to the Power of Something

1. Mar 6, 2009

### hurliehoo

1.

Hi could someone kindly give me a clue as to how I would go about calculating a matrix to the power of something?

eg A=[1 3^(1/2);
3^(1/2) 1]

what would A^12 be?

...
3.

Do I use P,P^(-1) and a diagonal matix somehow? I am really a beginner at matrices sorry!

2. Mar 6, 2009

### lurflurf

A^2=2A+2I
so
A^12=49920A+36544I

3. Mar 7, 2009

### hurliehoo

Sorry but that's not what I get when I check this with Matlab. I get :

A=
[1, 3^(1/2);
3^(1/2), 1]

B=A^12

B=
[14.74, 28.15;
28.15, 53.78]

But no idea how this is done with pencil and paper. Please enlighten me someone!

4. Mar 7, 2009

### gabbagabbahey

Matlab's answer is incorrect...are you sure you typed in in correctly?

If you don't like lurflurf's method (or don't follow it), you can of course start by diagonalizing A...If D=PAP-1 is diagonal, then A^12=(P-1DP)^12=(P-1DP)(P-1DP)(P-1DP)......(P-1DP)=P-1D12P since PP-1=I

5. Mar 8, 2009

### hurliehoo

I typed it in correctly I'm pretty sure of that. There was a scalar value in front which I multiplied with to give the above matrix. Oh well looks like it's wrong anyway. :(

Okay must admit I don't understand this. This is complex notation?

6. Mar 8, 2009

### HallsofIvy

No, there's nothing "complex" about it!

As lurflurf told you, A^2= 2A+ I. (He got that by finding its characteristic equation. Every matrix satisifies its own characteristic equation.)

So A^4= (A^2)^2= (2A+ 2I)^2= 4A^2+ 8A+ 4I= 4(2A+ 2I)+ 8A+ I= 16A+ 9I and, then, A^8= (A^4)^2= (16A+ 9I)^2= 256A^2+ 288A+ 81I= 256(2A+ 2I)+ 288A+ 81I= 800A+ 593I. Finally, A^12= (A^8)(A^4)= (800A+ 593I)(16A+ 9I)= 12800A^2+16688A+ 5337I= 12800(2A+ 2I)+ 16688A+ 5337I. See what that gives. Unless I have made an arithmetic error, that should give 49920A+ 36544I.

7. Mar 8, 2009

### lurflurf

I=a^0={{1,0},{0,1}}

8. Mar 8, 2009

### hurliehoo

Aaah alright starting to make a bit more sense now.