Let P=3/21/2−1/23/2,A=1    10    1 and Q=PAP′ then P′Q2015P is

P=cos⁡(π/6)sin⁡(π/6)−sin⁡(π/6)cos⁡(π/6)⇒ P′=cos⁡(π/6)−sin⁡(π/6)sin⁡(π/6)cos⁡(π/6)Since PP′=1    00    1⇒P′=P−1We have Q=PAP′=PAP−1⇒ Q2015=PAP−12015=PA2015P−1Thus, P′Q2015P=P−1PA2015P−1P=P−1PA2015P−1PNow, A=I+B where B=0    10    0Since, B2=O, we get Br=O∀r≥2Thus,  A2015=I+2015B=1201501