Use elementary methods to find a Nash equilibrium and the value for the game $A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 3 & 1 \end{bmatrix}$.
My thought:
There is no Nash Equilibrium, for whatever row that is first chosen the second player can't choose another row that best favors him/her.
Is this correct? What is the correct "elementary" procedure? I am also confused with the use of a single number, opposed to the paired number (x,y). What is the difference? What is the use of the paired numbers in the game theory matrix?
1) Since you are given only a matrix and you mention "value", one should assume that this is a zero-sum game. If you are used to bimatrix games, this is equivalent to $$A = \begin{bmatrix} 2,-2 & -1,+1 & 0,0 \\ -1,+1 & 3,-3 & 1,-1 \end{bmatrix}$$ where the sum of the payoffs of $P1$ and $P2$ is zero. Clearly, zero-sum games are a special case of bimatrix games (as the payoffs for these may not sum to zero).
2) This game has no Nash equilibria in pure strategies because it is never the case that at a given cell both players are playing a best response.
3) The celebrated Nash existence theorem states that every finite game has at least one equilibrium, possibly in mixed strategies. Hence, this game must have an equilibrium in mixed strategies. Given that this is a zero-sum game, this equilibrium may be bound using the minimax theorem. The equilibrium payoff for $P1$ is known as the value of the zero-sum game.