M 11 4 2 5 Find formulas for the entries of Mn where n is a

M = [11 -4 2 5] Find formulas for the entries of M^n, where n is a positive integer.

Solution

The characteristic polynomial of M mis det (M- I₂) = 0 or, ² -16+63 = 0 or (-7)( -9) = 0. Thus, the eigenvalues of M are ₁ = 9 and ₂ = 7. The eigenvectors of A corresponding to the eigenvalue 9 is the solution to the equation (A-9I₂) X = 0. To solve this equation, we will reduce to its RREF,as under, the matrix A-9I₂ =

2

2

-4

-4

Multiply the 1st row by ½

Add 4 times the 1st row to the 2nd row

Then the RREF of A-9I₂ is

1

1

0

0

Now, if X =(x,y)^T, then the equation (A-9I₂) X = 0 is equivalent to x+y = 0 or, x= -y.Then X =(-y,y)^T =y (-1,1)^T. Thus, the eigenvector of M corresponding to the eigenvalue 9 is (-1,1)^T.Similarly, the RREF of A-7I₂ is

1

1/2

0

0

Thus, the eigenvector of M corresponding to the eigenvalue 7 is (-1,2)^T.

Now, since M has 2 distinct linearly independent eigenvectors, it can be diagonalized as M = PDP^-1, where D =

9

0

0

7

The eigenvalues of M are the entries on the leading diagonal of D.

Also P =

-1

-1

1

2

The eigenvectors of M are the columns of P( in the same order).

Also, then P^-1 =

-2

-1

1

1

Then Mⁿ = (PDP^-1)ⁿ = PDⁿP^-1. Here, Dⁿ =

9ⁿ

0

0

7ⁿ

Then Mⁿ =

-7ⁿ+2*9ⁿ

-7ⁿ+9ⁿ

2*7ⁿ-2*9ⁿ

2*7ⁿ-9ⁿ

2	2
-4	-4