We have #A=((1,-1),(1,1))#.How to find #A^1995#?
The matrix #A# characteristic polynomial gives
#p(lambda)=lambda^2-2lambda+2# and #A# obeys it.
#A^2-2A+2I=0# or
#A^3-2A^2+2A = 0#
This obeys a recurrence relationship as in the difference equation
#a_(n+2)-2a_(n+1)+a_n=0# with solution
#a_n = A^n = C_1 2^(n/2)cos((n pi)/4)+C_2 2^(n/2)sin((n pi)/4)#
From #A# and #A^2# we can compute the #C_1, C_2# matrix components giving
#C_1 = ((1,0),(0,1))#
#C_2 = ((0,-1),(1,0))#
Then
#A^1995 = sqrt(2) xx 2^997(-1/sqrt(2)((1,0),(0,1))+ 1/sqrt2((0,-1),(1,0))) # or
#A^1995 = 2^997((-1,-1),(1,-1))#
