Let us recall that the famous fibonacci sequence: 0; 1; 1; 2; 3; 5; 8; 13; 21; : : : is de ned as follows: we put '0 = 0, '1 = 1 and de ne 'n+2 = 'n+1 + 'n: we want to nd a formula for 'n. to do this a) find a 2 2 matrix a such that 'n+2 'n+1 = a 'n+1 'n hint: combine the trivial equation 'n+1 = 'n+1 with the fibonacci relation 'n+2 = 'n+1 + 'n. b) diagonalize a and nd a formula for an. c) noticing that 'n+1 'n = an '1 '0 = an 1 0 nd a formula for 'n. (you will need to compute an inverse and perform multiplication here). d) show that the vector ('n+1='n; 1)t converges to an eigenvector of a. what do you think, is it a coincidence?