If A=100101010, then

A2=100101010100101010=100110101A3=100101011100110101=100201110A3−A2=0001−1101−1and A−I=0001−1101−1⇒ A3−A2=A−I                                (1)Now, det⁡An−I=det⁡(A−I)I+A+A2+…+An−1=det.(A−I)×det.I+A+A2+…+An−1=0From (1), A4 - A3 = A2 - A                         (2)Again from (1), A5 - A4 = A3 - A2 = A - IThus, if n is even, An - An-1 = A2 - A          (3)If n is odd, An - An-1 = A - I                        (4)Consider that n is even.∴ An−An−1=A2−A            [from(3)]and An−1−An−2=A−I            [from(4)]Adding these, we getAn−An−2=A2−I⇒ An=An−2+A2−I            =An−4+A2−I+A2−I            =An−6+A2−I+2A2−I                  ⋯ … …                   ⋯ ⋯ …            =A2+n−22A2−I∴ An=n2A2−n−22I∴ A200=100A2−99I                 =100100110101−99100010001=1001001010001