 Solution

Let $p$ be the row player's mixed strategy and $q$ the column player's mixed strategy. The fourth row is strictly dominated by the first. A useful candidate mixes the first two rows so that their first and third column payoffs agree. If the first row has probability $t$, these payoffs are $7t-2$ and $2-7t$, giving $t=2/7$ and common value zero.

To prove that these are optimal mixed strategies, use the two explicit mixed strategiesTheir payoff vectors satisfyThus the row player guarantees a payoff at least zero against every column mixed strategy, and the column player guarantees a payoff at most zero against every row mixed strategy. This matching pair proves that the value of the zero-sum game is zero.

These optimal mixed strategies are also unique. Since $Aq$ is strictly negative in rows three and four, any optimal row strategy gives them probability zero. Against columns one and three it must then satisfy both $7t-2\ge 0$ and $2-7t\ge 0$, so $t=2/7$. Similarly, $p^{T}A$ is strictly positive in column two, forcing an optimal column strategy to avoid that column; its constraints against rows one and two then force equal weights on columns one and three.