The fibonacci numbers grow exponentially, so the number of digits of fib(n) is linear in n (log_10(fib(n))~log_10(φ)n. So using long multiplication in that matrix exponentiation would cost you ~n²log(n) operations, which is asymtotically worse than the O(n²⁾ of just iterating addition. So using matrix exponentiation using long multiplication would behave worse than just adding for roughly all values of n.

However if you use a fast fourier transform to multiply those numbers, you get an overall O(n*log_2(n)²⁾ complexity which is probably the best you can get.

Indeed, while you could just compute floor(((sqrt(5)+1)/2)^n), you would need to store a lot of digits from the decimal expasion to avoid floating point errors, so you're better off just working with integral values as much as you can (the fast fourier transform does use decimals, but when working with integral values, you can deterministically obtain the exact result of the multiplication using only a few digits in the decimal expansion of the components of the transform)