I want to find the Nash-equilibrium of the following zero sum game. $$A=\begin{bmatrix}0&2&-1\\-2&0&3\\1&-3&0\end{bmatrix}$$
I used the Minimax Theorem.
$$min_{x \in X} max_{y \in Y} E(x,y)= min_{(u,x)} \{u; u \ge \sum^{n}_{i=1}a_{ij}x_i, \sum^{n}_{i=1}x_i=1, x_i \ge 0 \} $$
$$max_{y \in Y} min_{x \in X} E(x,y)=max_{(w,y)} \{w; w \le \sum^{m}_{j=1}a_{ij}y_j, \sum^{m}_{j=1}y_j=1, y_j \ge 0 \} $$
$$E(x,y)=x^TAy$$
I can get two linear programs
$$min \, u$$ $$-u-2x_2+x_3 \le 0 $$ $$-u+2x_1-3x_3 \le 0 $$ $$-u-x_1+3x_2 \le 0 $$ $$ x_1+x_2+x_3 =1 $$ $$ x_1,x_2,x_3 \ge 0 $$
and
$$max \, w $$ $$ w-2y_2+y_3 \le 0 $$ $$w+2y_1-3y_3 \le 0 $$ $$w-y_1+3y_2 \le 0 $$ $$ y_1+y_2+y_3 =1 $$ $$ y_1,y_2,y_3 \ge 0 $$
Then, how can I find the Nash-equilibrium?
Let one player's mixed strategy be $(p,q,r)$. Using the indifference condition for mixed NE, the other player's payoffs from each pure strategy must satisfy $$ \begin{aligned} -2q+r&=2p-3r\\ 2p-3r&=-p+3q\\ 1-p&=q+r \end{aligned} $$ Line 1 implies $$ 2r=p+q $$ Line 2 implies $$ p=q+r $$ Together with line 3, we can deduce $$ p=\frac12,\quad q=\frac16,\quad r=\frac13. $$ So this is the mixed NE with symmetric strategies.