1. Method & Example1
Method
Step1:

Player A's optimal strategies `=([[1,1]] xx P_(Adj))/([[1,1]] xx P_(Adj) xx [[1],[1]]) = [[p_1,p_2]]`

Step2:

Player B's optimal strategies `=([[1,1]] xx P_(Cof))/([[1,1]] xx P_(Adj) xx [[1],[1]]) = [[q_1,q_2]]`

Step3:

Value of the game = (Player A's optimal strategies) `xx` (Payoff matrix P) `xx` (Player B's optimal strategies)
`V = [[p_1,p_2]] xx P xx [[q_1],[q_2]]`

Example1
1. Find Solution of game theory problem using matrix method
_{Player A}\^{Player B}  B1  B2  B3  A1  1  7  2  A2  6  2  7  A3  5  1  6 
Solution: 1. Saddle point testing Players
   Player `B`       `B_1`  `B_2`  `B_3`    Player `A`  `A_1`   1  7  2   `A_2`   6  2  7   `A_3`   5  1  6  
We apply the maximin (minimax) principle to analyze the game.
   Player `B`       `B_1`  `B_2`  `B_3`   Row Minimum  Player `A`  `A_1`   1  7  2   `1`  `A_2`   (6)  [2]  7   `[2]`  `A_3`   5  1  6   `1`   Column Maximum   `(6)`  `7`  `7`   
Select minimum from the maximum of columns Column MiniMax = (6)
Select maximum from the minimum of rows Row MaxiMin = [2]
Here, Column MiniMax `!=` Row MaxiMin
`:.` This game has no saddle point.
2. Dominance rule to reduce the size of the payoff matrix Using dominance property
   Player `B`       `B_1`  `B_2`  `B_3`    Player `A`  `A_1`   1  7  2   `A_2`   6  2  7   `A_3`   5  1  6  
row3 `<=` row2, so remove row3
   Player `B`       `B_1`  `B_2`  `B_3`    Player `A`  `A_1`   1  7  2   `A_2`   6  2  7  
column3 `>=` column1, so remove column3
   Player `B`       `B_1`  `B_2`    Player `A`  `A_1`   1  7   `A_2`   6  2  
reduced matrix
   Player `B`       `B_1`  `B_2`    Player `A`  `A_1`   1  7   `A_2`   6  2  
For this reduced matrix, calculate `P_(Adj)` and `P_(Cof)`
`P_(Adj) = [[2,7],[6,1]]`
and `P_(Cof) = [[2,6],[7,1]]`
Player A's optimal strategies `=([[1,1]] xx P_(Adj))/([[1,1]] xx P_(Adj) xx [[1],[1]])`
`=([[1,1]][[2,7],[6,1]])/([[1,1]][[2,7],[6,1]][[1],[1]])`
`=([[4,6]])/(10)`
`=[[2/5,3/5]]`
`p_1=2/5` and `p_2=3/5`, where `p_1` and `p_2` represent the probabilities of player A's, using his strategies `A_1` and `A_2` respectively.
Similarly, Player B's optimal strategies `=([[1,1]] xx P_(Cof))/([[1,1]] xx P_(Adj) xx [[1],[1]])`
`=([[1,1]][[2,6],[7,1]])/([[1,1]][[2,7],[6,1]][[1],[1]])`
`=([[5,5]])/(10)`
`=[[1/2,1/2]]`
`q_1=1/2` and `q_2=1/2`, where `q_1` and `q_2` represent the probabilities of player B's, using his strategies `B_1` and `B_2` respectively.
Hence, Value of the game `V` = (Player A's optimal strategies) `xx` (Payoff matrix `P_(ij)`) `xx` (Player B's optimal strategies)
`V=[[2/5,3/5]][[1,7],[6,2]][[1/2],[1/2]]=4`
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
