daca iinmultim A *A obtinem
A²= (6 6) =A
(-5 -5)
pt ca 6*6-6*5=6*(1=6
si 6*(-5) +(-5)(-5) =-30+25=-5
presupunem A^n=A
verificam pt n=1 (ipoteeza) si pt n=2 (mai sus)
presupunem adevarat pt n=n
atunci
A^(n+1)=A^n * A=A*A=A²=A
Pn⇒P(n+1) deci demonstrat prin inductie completa ca A^n=A,∀n∈N*
atunci A^2012=A^2011=A
si relatia ceruta devine A-A=O2 , adevarat
