Pure Maths---Matrix3

a) if L is an eigen value of A
ie. there exist a V, such that
AV=LV

since P is invertable, let Q be P^-1 (easy for me)
B=QAP => QA=BQ
thus,

AV=LV
QAV=LQV
BQV=LQV

thus L is also eigen value of B with eigen vector QV

b)
we notice that k1, k2, k3 are eigen value of B thus, by a) is it also eigen value of A
by standard method, Det[ A- I L]=0
we found the eigen value of A are -2,-1,1
thus,
k1=-2, k2=-1, k3=1

since AP=PB
we treat P = [ v1 v2 v3] where v1 v2 v3 are column vector (vertical)
v1= [ 1 a b]
v2= [ c 1 d]
v3= [e f 1]

thus A[v1 v2 v3]= [k1v1 k2v2 k3v3]
separate
Avi=kivi, where i=1,2,3
thus vi are the egenvector for A with corresponding eigenvalue ki
by bi)
v1 = {1, -1, -1}
v2 = {-2, 1, 0}
v3={0, 0, 1}

therefore, a=-1, b=-1, c=-2, d=e=f=0

P = { 1 -2 0; -1 1 0; -1 0 1}

iii) A=PBQ
A^2n = PB^2nQ since PQ=QP=1
B^2n = [ (-2)^2n 0 0; 0 (-1)^2n 0; 0 0 1]
= [ 4^n 0 0; 0 1 0; 0 0 1]
found Q,
put it into PB^2nQ, after grouping term, will give the result. again, i just give you a guide line, try to do it yourself.