Method

Step-1:	Player A's optimal strategies `=([[1,1]] xx P_(Adj))/([[1,1]] xx P_(Adj) xx [[1],[1]]) = [[p_1,p_2]]`
Step-2:	Player B's optimal strategies `=([[1,1]] xx P_(Cof))/([[1,1]] xx P_(Adj) xx [[1],[1]]) = [[q_1,q_2]]`
Step-3:	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]]`
	Here `P_(Cof)` = Cofactor matrix of P and `P_(Adj)` = Adjoint matrix of P

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Example-1
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.

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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

row-3 `<=` row-2, so remove row-3

			Player `B`
			`B_1`	`B_2`	`B_3`
Player `A`	`A_1`		1	7	2
	`A_2`		6	2	7

column-3 `>=` column-1, so remove column-3

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_1`		1	7
	`A_2`		6	2

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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`

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