If Rolle's theorem holds for the function $$f(x) = x^3 - ax^2 + bx - 4$$, $$x \in [1, 2]$$ with $$f'\left(\frac{4}{3}\right) = 0$$, then ordered pair $$(a, b)$$ is equal to:

Given $$f(x) = x^3 - ax^2 + bx - 4$$ on $$[1, 2]$$ with Rolle's theorem applicable, we need $$f(1) = f(2)$$ and $$f'\left(\frac{4}{3}\right) = 0$$.

Computing $$f(1) = 1 - a + b - 4 = -3 - a + b$$ and $$f(2) = 8 - 4a + 2b - 4 = 4 - 4a + 2b$$.

Setting $$f(1) = f(2)$$: $$-3 - a + b = 4 - 4a + 2b$$, which simplifies to $$3a - b = 7$$ ... (i).

Now $$f'(x) = 3x^2 - 2ax + b$$. Setting $$f'\left(\frac{4}{3}\right) = 0$$: $$3 \cdot \frac{16}{9} - 2a \cdot \frac{4}{3} + b = 0$$, which gives $$\frac{16}{3} - \frac{8a}{3} + b = 0$$. Multiplying by 3: $$16 - 8a + 3b = 0$$ ... (ii).

From (i): $$b = 3a - 7$$. Substituting into (ii): $$16 - 8a + 3(3a - 7) = 0$$, which gives $$16 - 8a + 9a - 21 = 0$$, so $$a - 5 = 0$$, hence $$a = 5$$.

Then $$b = 3(5) - 7 = 8$$.

Therefore, the ordered pair $$(a, b) = (5, 8)$$.