I am asked to give an example of a $3\times 3$ matrix game where PI has a totally mixed optimal strategy, but also has an optimal strategy which is not totally mixed.
So I know that $S_1=(x_1,x_2,x_3)$ such that $x_1>0,x_2>0,x_3>0, x_3=1-x_1-x_2$ and that $S_2=(y_1,y_2,0)$ such that $y_1>0,y_2>0, y_2=1-y_1$
I guess I am just not sure how to attack this problem, aside from just generating random $3\times 3$ matrix games and seeing if they work...
I assume as stated in the question that you mean matrix game as in two-player zero-sum game (with a non-zero sum, bimatrix game, the next paragraph is not true).
This is not possible unless the game is degenerate (meaning that there exists a mixed strategy $p$ that uses $k$ pure strategies with positive probability that has strictly more that $k$ best responses to it). If it is true then it must be the case that there are infinitely many equilibria.
A trivial solution is the following:
$$\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) $$
since now all mixed strategy profiles are equilibria.
To create a less trivial example, I would start with a hide-and-seek game which has a unique totally mixed equilibrium (even though it is already degenerate) and then alter it. A $3\times3$ hide-and-seek game is:
$$\left( \begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array} \right) $$
So here the column player wins if he picks the same location as the row player, otherwise the row (hider) player wins. We can actually simplify and just make it $(-I,I)$ (so now they draw if the row player is not found), which also has a unique completely mixed equilibrium:
$$\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right) $$
Finally to make it have more than one equilibrium, let's add a second pure best response against row 1 , namely column 2, to give:
$$\left( \begin{array}{ccc} -1 & -1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right) $$
Now we have the following equilibria: Player 1 players $(p, 1-p,0.5)$ where $p \in [0,0.5]$ and player 2 plays $(0, 0.5, 0.5)$. Intuitively, now the column player never plays column 1, always preferring column 2, but the row player can still mix in all three columns since when the column player has zero probability on column 1, rows 1 and 2 are payoff equivalent and can be played individually or mixed.
BTW, here is the output for this game as found by my game solver http://banach.lse.ac.uk/ (or you could use our newer http://www.gametheoryexplorer.org/).
3 x 3 Payoff matrix A: 1 1 0 0 1 0 0 0 1 3 x 3 Payoff matrix B: -1 -1 0 0 -1 0 0 0 -1 EE = Extreme Equilibrium, EP = Expected Payoff Decimal Output EE 1 P1: (1) 0.5 0.0 0.5 EP= 0.5 P2: (1) 0.5 0.0 0.5 EP= -0.5 EE 2 P1: (1) 0.5 0.0 0.5 EP= 0.5 P2: (2) 0.0 0.5 0.5 EP= -0.5 Rational Output EE 1 P1: (1) 1/2 0 1/2 EP= 1/2 P2: (1) 1/2 0 1/2 EP= -1/2 EE 2 P1: (1) 1/2 0 1/2 EP= 1/2 P2: (2) 0 1/2 1/2 EP= -1/2 Connected component 1: {1} x {1, 2}