If n is a positive integer then 14 40 4 14n is 14 14 2 2Solu

If n is a positive integer, then [14 -40 4 -14]^n is [14 14 2 2]

Solution

The characteristic equation of A is det (A- I₂) = 0 or, ² -36= 0 or,( +6)( -6) = 0. Thus, A has 2 distinct eigenvalues, ₁ = -6 and ₂ = 6.

The eigenvector of A corresponding to the eigenvalue is solution to the equation (A- I₂)X = 0. Thus, the eigenvectors of A corresponding to the eigenvalue -6 are solutions to the equation (A+6I₂)X = 0.We will reduce A+6I₂ to its RREF as under:

Multiply the 1st row by 1/20

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

Then the RREF of A is

1

-2

0

0

  If X = (x,y)^T , then the above equation is equivalent to x-2y = 0 or, x = 2y so that X = (2y,y)^T = y(2,1)^T. Thus, the eigenvector of A corresponding to the eigenvalue -6 is (2,1)^T. Similarly, the eigenvector of A corresponding to the eigenvalue 6 is (5,1)^T. Now, let D =

-6

0

0

6

and P =

2

5

1

1

Then P^-1 =

-1/3

5/3

1/3

-2/3

Further, A = PDP^-1 and Aⁿ = PDⁿP^-1. Now, Dⁿ =

(-6)ⁿ

0

0

6ⁿ

so that Aⁿ =

5*2ⁿ*3^n-1 –(-1)ⁿ*2ⁿ⁺¹*3^n-1

-5*2ⁿ⁺¹*3^n-1+5*(-1)^n*2ⁿ⁺¹*3^n-1

-(-2)ⁿ*3^n-1+ 2²*3^n-1

5(-2)ⁿ*3^n-1- 2ⁿ⁺¹*3^n-1

1	-2
0	0