Given this plaintext: 101010101010, and the block is 3, with a IV = 000 and the key as k = |1 2 3| |2 1 3| What are the required steps to get the same result like these in ECB mode: c1 = Ek(m1) = 011, c2 = Ek(m2) = 100, c3 = Ek(m3) = 011 et c4 = Ek(m4) = 100. c = c1||c1||c1||c1 = 011100011100.
Note: It is not a homework/ test, I just making some preparation.