You need to use Cayley-Hamilton theorem, hence, you need to evaluate the characteristic equation of the given matrix A, such that:

`lambda^2 - Tr(A)*lambda + Delta A*I_2 = O_2`

`Tr(A) = 2 - 8 =-6`

`Delta A= 2*(-8) - 5*(-5) => Delta A = -16 + 25 => Delta A = 9`

`lambda^2 + 6lambda + 9 = 0 => (lambda + 3)^2 = 0 => lambda_1 = lambda_2 = -3`

You may evaluate the nth power of matrix A, such that:

`A^n = (-3)^n*B + (-3)^n*n*C`

Selecting n=0 and n=1 yields:

`A^0 = (-3)^0*B + (-3)^0*0*C`

`A^1 = (-3)^1*B + (-3)^1*1*C`

`{(B = I_2),(-3(B+C) = A):}`

Considering `C = ((c_1, c_2),(c_3, c_4))` yields:

`C + B = C + I_2 = ((c_1 ,c_2),(c_3, c_4)) + ((1, 0),(0, 1))`

`C + B = ((c_1 + 1, c_2),(c_3 ,c_4 + 1))`

`((c_1 + 1, c_2),(c_3, c_4 + 1)) = ((2, 5),(-5, -8)) => {(c_1 + 1 = 2),(c_2 = 5),(c_3 = -5), (c_4 + 1 = -8):}`

`{(c_1 = 1),(c_2 = 5),(c_3 = -5), (c_4 = -9):}`

`C = ((1, 5),(-5, -9))`

`n*C = ((n ,5n),(-5n, -9n))`

`A^n = (-3)^n*I_2 + (-3)^n*((n, 5n),(-5n ,-9n))`

**Hence, evaluating the matrix A, using Cayley-Hamilton theorem, yields **`A^n = (-3)^n*I_2 + (-3)^n*((n,5n),(-5n ,-9n)).`