I have a payoff matrix :
\begin{pmatrix} 5 & -2 & -5 \\ -2 & 3 & 2 \\ -3 & 6 & 2 \\ 4 & -8 & -6 \\ \end{pmatrix}
I need to find the optimal stategies. I first tried writing this as a Linear Program by making all the values positive (adding 8 to each entry) and applying the simplex algorithm. However, the number are very horrible... Is there a another way to find the strategies?
If the numbers appearing in the Simplex algorithm are very horrible, I doubt if there's any way of finding the solution that will eliminate the possibility of having at some stage to deal with those very horrible numbers.
In the revised version of the game, as now amended, however, player I's fourth strategy is strictly dominated by his first, so you can restrict your attention to the game with matrix $$ \begin{pmatrix} 5&-2&-5\\ -2&3&2\\ -3&6&2 \end{pmatrix}\ . $$ One thing you could try is first to assume that one of the player's strategies assigns positive probabilities to all of his pure strategies in this game, and if this doesn't work, try solving the nine games with each of the $2\times 2$ square submatrices of the above matrix. A solution for one of these games will be optimal in the original if any pure strategy of either player not included in the mix is no better for him or her than those that are so included.
The problem with this method is that it's really just a less systematic way of doing what the Simplex method effectively does anyway. While it's possible that you could stumble across an optimal pair of strategies without having to deal with any of the submatrices that give rise to the very horrible numbers, there's no guarantee of that.
As it happens, if you don't add $8$ to every entry of the matrix, but instead leave it as it is, none of the numbers that turn up seem to me to be very horrible. If player II's optimal strategy $\ \left[p_1^*, p_2^*, p_3^*\right]\ $ has $\ p_i^*>0\ $ for every $\ i\ $ the player I's, $\ \left[q_1^*, q_2^*, q_3^*\right]\ $, say, must satisfy the equation $$ \begin{pmatrix} q_1^*&q_2^*&q_3^* \end{pmatrix} \begin{pmatrix} 5&-2&-5\\ -2&3&2\\ -3&6&2 \end{pmatrix} = v\begin{pmatrix} 1&1&1 \end{pmatrix}\ $$ where $\ v\ $ is the value of the game. This gives \begin{eqnarray} \begin{pmatrix} q_1^*&q_2^*&q_3^* \end{pmatrix}&=&v\begin{pmatrix} 1&1&1 \end{pmatrix} \begin{pmatrix} 5&-2&-5\\ -2&3&2\\ -3&6&2 \end{pmatrix}^{-1}\\ &=& v\begin{pmatrix} 1&1&1 \end{pmatrix} \begin{pmatrix} \frac{6}{11}&\frac{26}{11}&-[\\ \frac{2}{11}&\frac{5}{11}&0\\ \frac{3}{11}&\frac{24}{11}&-1 \end{pmatrix}\\ &=& v \begin{pmatrix} 1&5&-2 \end{pmatrix}\ , \end{eqnarray} and it's impossible for any scalar multiple of $\ \begin{pmatrix} 1&5&-2 \end{pmatrix}\ $ to have all positive entries.
I have to admit, I didn't try going through the rigmarole of solving the games with all nine $\ 2\times 2\ $ submatrices, but instead submitted the full game matrix to this online matrix game solver. In the solution it returned player I only chooses his first two pure strategies with positive probabilities, and player II only chooses his first and third. The $2\times 2$ matrix game in which the players are limited to choosing those pure strategies is $$ \begin{pmatrix} 5&-5\\ -2&2 \end{pmatrix}\ . $$ The optimal mixed strategies in this game are $\ \left[\frac{2}{7},\frac{5}{7}\right]\ $ for player I, and $\ \left[\frac{1}{2},\frac{1}{2}\right]\ $ for player II. Converting these to mixed strategies, $\ \left[\frac{2}{7},\frac{5}{7},0,0\,\right]\ $ and $\left[\frac{1}{2},0,\frac{1}{2}\right]\ $, for the original game, we find that $$ \begin{pmatrix} \frac{2}{7} & \frac{5}{7} & 0 & 0 \end{pmatrix} \begin{pmatrix} 5 & -2 & -5\\ -2&3&2\\ -3&6&2\\ 4&-8&-6 \end{pmatrix} = \begin{pmatrix} 0 & \frac{11}{7} & 0 \end{pmatrix}\ , $$ and $$ \begin{pmatrix} 5 & -2 & -5\\ -2&3&2\\ -3&6&2\\ 4&-8&-6 \end{pmatrix} \begin{pmatrix} \frac{1}{2} \\ 0 \\ \frac{1}{2} \end{pmatrix} = \begin{pmatrix} 0\\ 0 \\ -\frac{1}{2} \\ -1 \end{pmatrix}\ , $$ so these strategies are indeed optimal, and the value of the game is $0$ .