Would have to look at the latest source code 
The algorithms they do use (like Toledo’s recursive LU decomposition) are known to be implemented and tuned in ways that make them run fast on both small and huge matrices by optimizing cache accesses/locality of reference, parallel processing, fast matrix multiplication where appropriate, and whatever else they can think of, so, whatever they have implemented, I trust it is meant to be as fast as possible (that they know how to do) on real hardware.
If the centers of the balls are 10 meters apart when they leave Nina’s and Pascal’s hands, they meet when the centers are 10 centimeters or 0.1 meter apart, so isn’t the distance each travels (10 - 0.1) / 2 = 4.95 meters?
These sort of problems always bothered me because of the possible ways we can interpret “standing 10 meters apart” and “throw suddenly”. If I am 10 meters away from you but my arms are nearly a meter long, when I release the ball after accelerating it to 1 m/s it is going to be less than 10 meters away from you at the point of release.
The radius of each individual ball is 10 cm, so that adds up.
It is unsurprisingly often easy to nitpick the types of toy word problems described in this thread. (Especially if one is in the railroad industry.) My interpretation was that “throw suddenly” and “10 meters away” means that the centers start out 10 m apart with each one moving at 1 m/s.
A quaternion leaves Los Angeles headed for Denver traveling at O(n[sup]3[/sup]) speed. At the same time, an matrix leaves Denver headed for Los Angeles at O(3[sup]n[/sup]) speed. At what cross product along the way do their eigenvectors meet?
Divide by cucumber error.
I don’t think so. I think they meet when their centers are 20 cm apart. Remember the balls have a radius of 10 cm, not a diameter of 10 cm.
I am standing by my original answer of 4.9 m in 4.9 seconds.