We have to find the equilibria of the zero sum game specified by the following matrix: \begin{matrix} \\ & A & B & C \\ T &-2 & 10 & 4 \\ M & 7 & 8 & 7 \\ B & 8 & 1 & 4 \end{matrix}
Clearly, a saddle point is - 1 plays M, 2 plays C, which gives a value of 7. My question is, are there any other equilibria of this game? Also, in general, if a zero sum game has saddle points - i.e. pure strategy equilibria - can it still have additional mixed strategy equilibria?
In general - yes. For example, take the regular $2\times 2$ matching pennies and add an action to each player, where all payoffs in the added row and column are $0.5$. This game has the previous mixed equilibrium and some new ones in which one or both players play their new action.
Here, we can look for more equilibria using the following theorem for zero-sum games: if $(s_1,s_2)$ is an equilibrium and $(b_1,b_2)$ is another equilibrium, then also $(s_1,b_2)$ and $(b_1,s_2)$ are equilibria. The idea is that in a zero-sum game, an action is "optimal" and then can be played against any action of the opponent.
For example here, we deduced that $C$ is optimal for 2. This means that if there is another equilibrium where player 1 mixes, he can play it against $C$ and get the same result. But any mixture against $C$ is sub-optimal, so Player 1 has no mixed optimal strategy.
You can do the same exercise for player 2 and see that he can mix $A$ and $C$, but he should play $A$ with low enough probability so that $M$ will still be the best response of Player 1. (lower than $0.75$).
To conclude here there are infinitely many equilibria of the form: Player 1 plays $M$, player 2 mixes $A$ w.p. $x\leq 0.75$ and $C$ w.p. $1-x$.