docs: public double[] solveLInplace(double[] X) {
      final int n = L.length;
      X[0] /= L[0][0]; // First iteration, simplified.
      for(int k = 1; k < n; k++) {
        final double[] Lk = L[k];
        for(int i = 0; i < k; i++) {
          X[k] -= X[i] * Lk[i];
        }
        X[k] /= Lk[k];
      }
      return X;
    }