數學問題(d,in,matrix)

Using the result in (b).
Let x = A and n = 2012, we get
A^2012 = f(A)g(A) + pA^2 + qA + r
where
p = [(-1)^2012 + 2(2012) - 1]/4 = 2011/4
q = [1 - (-1)^2012]/2 = 0
r = [(-1)^2012 - 2(2012) + 3]/4 = -1005
Also f(A) = - (A + I)(A - I)^2 = 0 ( zero matrix). (Detail calculation not stated here, you can verify).
so A^2012 = 0 g(A) + (2011/4)A^2 - 1005A
= (2011/4)A^2 - 1005A.
Please verify the answer by yourself.


2011-10-31 08:34:03 補充：
Correction : p should be = 1006, sorry for the mistake. So A^2012 = 1006A^2 - 1005A.

2011-10-31 08:45:08 補充：
Correction : Again a mistake, A^2012 should be = 1006A^2 - 1005I, not 1005A. Sorry.

2011-10-31 08:48:03 補充：
Note : To be exact, A^2012 = f(A)g(A) + pA^2 + qA + rI.

可唔可以答埋17同47,PLZ
同埋,X可以係real number又可以係matrix?