Payoff matrix with a specific form I am very stuck on this question:
Suppose that $b \in \mathbb{R}^m$, $c \in \mathbb{R}^n$, $A$ is a $m \times n$ real matrix, and all components of $A$, $b$ and $c$ are positive. Consider the two-person zero-sum game in which each player has $m+n+1$ pure strategies and the payoff matrix is
\begin{equation}  M= \Bigg( \begin{array}{c:c} 0 & A& -b \\ \hdashline
-A^{T} & 0 & c  \\ \hdashline b^{T} & -c^{T} & 0 \end{array} \Bigg). \end{equation}
There are $3$ parts:
$1.$  Find the equilibrium payoff for player 1 (row player).
$2.$ If both players have the same optimal mixed strategy in the game being 
\begin{equation}
 \pi^{T} = (p_1, p_2, \ldots, p_m, q_1, \ldots, q_n, r),
\end{equation}
prove that $r \neq 0$.
$3.$ Explain how to find from $\pi$ an optimal solution to the linear program
\begin{equation}
 \{ \text{ maximize } c^{T} y \, : \, Ay \leq b, y\geq 0\}.
\end{equation}
 A: For 1. Use the following

Proposition 3.4 If the payoff matrix for a two-player zero-sum game is skew-symmetric,
i.e. $M^T$ = −M, then the value of the game is $0$ and the optimal mixed strategies for both
players are the same. The game is then said to be symmetric.

that you can find in this link or in any textbook.
For 2. Since $π$ is optimal and the game is $0$ sum we know that $π$ guarantees the value of the game (i.e. $0$) for player I against any strategy of player II (we will use also in part 3. below). Assume know that $r=0$, then $π$ against the first strategy of player II yields a payoff $$π^TMe_1=p^T0+q^T\left(-A^T_{\cdot 1}\right)+0\cdot c_1=-q^TA_{\cdot 1}$$ where with $A^T_{\cdot 1}$ we denote the first column of the matrix $A^T$. Now this payoff is negative since all the elements of $-A$ are negative unless $q$ is the zero vector. But if $q$ is the zero vector then the payoff of $π$ against the last strategy of Player II is equal to $$π^ΤΜe_{m+n+1}=-p^Tb+0<0$$ which is a contradiction. So $r>0$. Similarly we can show more, i.e. that $p,q$ are not the zero vectors.
For 3. Use that $π$ guarantees to Player I his security level against any strategy of Player II, i.e. that $π^ΤΜe_j\ge 0$ for every $j=1,2,\ldots, m+n+1$. This gives the set of inequalities $$\begin{cases}-q^TA^T+rb^T\ge0\\p^TA-rc^T\ge0\\-p^Tb+q^Tc\ge0\end{cases} \implies \begin{cases}\dfrac{q^T}{r}A^T\le b^T\\\dfrac{p^T}{r}A\ge c^T\\\dfrac{p^T}{r}b\le \dfrac{q^T}{r}c\end{cases}$$ The first inequality gives you that the vector $\dfrac{q^T}{r}$ is primal feasible, the second inequality that the vector $\dfrac{p^T}{r}$ is dual feasible and the third that the value of the objective function of the primal LP for the primal feasible vector $\dfrac{q^T}{r}$ is greater or equal than the value of the objective function of the dual LP for the dual feasible vector $\dfrac{p^T}{r}$. Due to strong duality (actually weak duality suffices) this yields optimality of the solutions.
