Here, $$z = {{ - 1 + i\sqrt 3 } \over 2} = \omega $$

$$\because$$ $$P = \left[ {\matrix{ {{{( - \omega )}^r}} & {{\omega ^{2s}}} \cr {{\omega ^{2s}}} & {{\omega ^r}} \cr } } \right]$$

$$ = \left[ {\matrix{ {{\omega ^{2r}} + {\omega ^{4s}}} & {{\omega ^{r + 2s}}[{{( - 1)}^r} + 1]} \cr {{\omega ^{r + 2s}}[{{( - 1)}^r} + 1]} & {{\omega ^{4s}} + {\omega ^{2r}}} \cr } } \right]$$

Given, $${P^2} = - I$$

$$\therefore$$ $${\omega ^{2r}} + {\omega ^{4s}} = - 1$$

and $${\omega ^{r + 2s}}[{( - 1)^r} + 1] = 0$$

Since, r $$\in$$ {1, 2, 3} and ($$-$$1)^r + 1 = 0

$$\Rightarrow$$ r = {1, 3}

Also, $${\omega ^{2r}} + {\omega ^{4s}} = - 1$$

If r = 1, then $${\omega ^2} + {\omega ^{4s}} = - 1$$

which is only possible, when s = 1.

As, $${\omega ^2} + {\omega ^4} = - 1$$

$$\therefore$$ r = 1, s = 1

Again, if r = 3, then

$${\omega ^6} + {\omega ^{4s}} = - 1$$

$$ \Rightarrow {\omega ^{4s}} = - 2$$ [never possible]

$$\therefore$$ r $$\ne$$ 3

$$\Rightarrow$$ (r, s) = (1, 1) is the only solution.

Hence, the total number of ordered pairs is 1.