SOLUTION: Let M=[4 4 ; −2 10]. Find formulas for the entries of M^n, where n is a positive integer.

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Question 1161347: Let M=[4 4 ; −2 10].
Find formulas for the entries of M^n, where n is a positive integer.

Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
We diagnalize the matrix:

matrix%281%2C3%2CM%2C%22%22=%22%22%2C%0D%0A%28matrix%282%2C2%2C4%2C4%2C-2%2C10%29%29%29

We find the eigenvalues

abs%28matrix%282%2C2%2C4-lambda%2C4%2C-2%2C10-lambda%29%29%29%22%22=%22%220

%284-lambda%29%2810-lambda%29-%28-2%29%284%29=0

40-14lambda%2Blambda%5E2%2B8=0

lambda%5E2-14lambda%2B48=0

%28lambda-6%29%28lambda-8%29=0

λ-6=0;  λ-8=0
  λ=6;    λ=8

by writing it as 

matrix%281%2C3%2CM%2C%22%22=%22%22%2CPDP%5E%28-1%29%29

where D is the diagonal matrix with the two eigenvalues on the 
main diagonal:

D+=+%28matrix%282%2C2%2C6%2C0%2C0%2C8%29%29

and the matrix P is 

P=%28matrix%281%2C2%2CV%5B1%5D%2CV%5B2%5D%29%29

where the V's are the two column eigenvectors for the two eigenvalues

We find V1 which is the eigengvector for the eigenvalue λ=6.

We find solutions for

%28M-6I%29X=0





-2x%5B1%5D%2B4x%5B2%5D=0
Divide thru by -2
x%5B1%5D-2x%5B2%5D=0
x%5B1%5D=2x%5B2%5D

We can take x1=1 and x1=1

So 

v%5B1%5D=%28matrix%282%2C1%2C2%2C1%29%29

Now we do the same for the other eigenvalue

---

We find solutions for

%28M-8I%29X=0





-4x%5B1%5D%2B4x%5B2%5D=0
Divide thru by -4
x%5B1%5D-x%5B2%5D=0
x%5B1%5D=x%5B2%5D

We can take x1=1 and x2=1

So 

v%5B1%5D=%28matrix%282%2C1%2C1%2C1%29%29

So

P=%28matrix%282%2C2%2C2%2C1%2C1%2C1%29%29

And since the determinant of P is 1, to find P-1 we only
need to swap the elements on the the main diagonal and change the
signs of the other two elements"

P%5E%28-1%29=%28matrix%282%2C2%2C1%2C-1%2C1%2C2%29%29

Then 

M=PDP%5E%28-1%29

So



to n factors



M%5En+=+P%2AD%5En%2AP%5E%28-1%29

Any power of a diagonal matrix is the matrix whose elements are that
power of the elements, so we have the final answer as:



Edwin