Here is the matrix:
[ [ 0, 1 ] * [a, b] = [b, a + b]
[ 1, 1 ] ]
Thus ‘[0, 1; 1, 1]^n * [1, 1]’ computes Fibonacci numbers ‘n+1’ and ‘n+2’. Here's one program that does the job:
C-x ( ' [0, 1; 1, 1] ^ ($-1) * [1, 1] <RET> v u <DEL> C-x )
This program is quite efficient because Calc knows how to raise a matrix (or other value) to the power ‘n’ in only ‘log(n,2)’ steps. For example, this program can compute the 1000th Fibonacci number (a 209-digit integer!) in about 10 steps; even though the Z < ... Z > solution had much simpler steps, it would have required so many steps that it would not have been practical.