If n is a positive integer then 2 4 120 2n is SolutionThe ch

If n is a positive integer, then [2 -4 -120 -2]^n is

Solution

The characteristic equation of A is det (A- I₂) = 0 or, ² -484= 0 or,( +22)( -22) = 0. Thus, A has 2 distinct eigenvalues, ₁ = -22 and ₂ = 22.

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 -22 are solutions to the equation (A+22I₂)X = 0. We will reduce A+22I₂ to its RREF as under:

Multiply the 1st row by 1/24      

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

Then the RREF of A is

1

-5

0

0

If X = (x,y)T , then the above equation is equivalent to x-5y = 0 or, x = 5y so that X = (5y,y)T = y(5,1)T. Thus, the eigenvector of A corresponding to the eigenvalue -22 is (5,1)T. Similarly, the eigenvector of A corresponding to the eigenvalue 22 is (-6,1)T. Now, let D =

-22

0

0

22

and P =

5

-6

1

1

Then P^-1 =

1/11

6/11

-1/11

5/11

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

(-22)ⁿ

0

0

22ⁿ

so that Aⁿ =

1/11(3*2ⁿ-11ⁿ)+5(-2)ⁿ*11^n-1

5/11(-3*2ⁿ+11ⁿ) +15(-1)ⁿ*2ⁿ⁺¹*11^n-1

(-2)ⁿ*11^n-1 -2ⁿ*11^n-1

5*2ⁿ*11^n-1+3(-1)ⁿ*2ⁿ⁺¹*11^n-1

1	-5
0	0