I am trying to solve the following $3\times4$ game: \begin{array}{c|rrrr} & A & B & C & D \\\hline X & -3 & 5 & 2 & -1 \\ Y & 4 & -1 & 1 & -3 \\ Z & -1 & -3 & -3 & 0 \end{array} I know that the optimal (mixed) strategy for the columns is $$\left(\frac 15A,\ 0\cdot B,\ \frac15C,\ \frac 35D\right)$$
How do I go about finding the optimal (mixed) strategy for the rows? I know how to do this using linear programming but I am looking for a technique that will be quicker. I also know how to do this for $2\times2 $ matrices but I am not sure how to extend this to $3\times3 $ matrices.
I suspect the subgame that needs to be realised is: \begin{array}{c|rrrr} & A & C & D \\\hline X & -3 & 2 & -1 \\ Y & 4 & 1 & -3 \\ Z & -1 & -3 & 0 \end{array}
Any help would be gratefully received.
Up to obtaining the reduced $3\times 3$ matrix, you did fine. Now, the optimal strategy of the rows player is an equalizing strategy, giving him the value of the game against any strategy of the opponent (column player). The value $v$ of the game can be calculated from the optimal strategy of the column player, if we pair this optimal strategy against any strategy of the rows player. So, for example against strategy $X$ $$v=-3\frac15+2\frac15-\frac35=-\frac45 $$ So, you need to solve \begin{cases}-3x+4y-\phantom{3}z&=-\frac45\\\phantom{-}2x+\phantom{3}y-3z&=-\frac45\\-\phantom{2}x-3y+0z&=-\frac45\\\phantom{-2}x+\phantom{3}y+\phantom{3}z&=1\end{cases}
Just to compare your result, by standard algebraic operations you can find $$(x,y,z)=\frac1{55}\left(17,\ 9,\ 29\right)$$