Matrices aren't the most intuitive way to think about that technique. Instead, it's about using the golden ratio, φ. Just like in Binet's formula, you want to raise φ to the power of the fibonacci number you want. BUT, don't use an approximation of phi. Instead, work it out directly with symbolic values of the form a + bφ. And then you don't need the rest of the formula: the nth fibonacci number will be what I called "b"; that is, the coefficient on φ.

To multiply a + bφ by c + dφ, you can just FOIL it out, and get ac + (bc + ad)φ + bdφ^2. But one interesting property of the golden ratio is that φ^2 = φ + 1. So you can substitute that to get rid of the φ^2, and get another expression of the form e + fφ. The punchline: you can do all of this math with a symbolic φ, using nothing but pairs of integers: a constant, and a coefficient on φ.

Now you want to raise this to the nth power. You can do that in O(log n) steps as follows: if n is 1, then you're done. If n is odd, then φ^n is just φ * φ^(n - 1). But if n is even, then you compute φ^(n / 2), and just square it. Because you half your even numbers, this is very fast: you can compute the millionth power of phi with fewer than 40 multiplications.

There's some python code for this at https://replit.com/@cdsmithus/fib if you're interested.