Explanation Given Information: The provided column matrix is C = [ 1 3 1 3 1 3 ] T . Formula used: The expected payoff e of the game is product of the matrices in the order RPC , where R is the row matrix, P is the payoff Matrix and C is the column matrix. The optimal strategy of the other player is the row matrix with highest payoff to the player where the other player uses the strategy C = [ a b c ] T . For matrices A with dimension m × n and B with dimension k × l , the product AB is defined if the number of columns in A equal number of columns in B , n = k . For matrices A with dimension m × n and B with dimension n × k , the product AB is the matrix of dimension m × k and i j t h entry of AB is the sum of product of corresponding entries of row i of A and column j of B . Calculation: Consider the payoff matrix P = [ 0 − 1 5 2 − 2 4 0 3 0 1 0 − 5 ] , and the column matrix C = [ 1 3 1 3 1 3 ] T . Take the matrix C = [ 1 3 1 3 1 3 ] T . This contains 1 row and 4 columns along with transpose sign. Transpose interchange the rows with columns and vice versa. Thus, matrix C is rewritten as, C = [ 1 3 1 3 1 3 ] T = [ 1 3 1 3 1 3 ] Let R = [ x y z t ] be the row matrix for the other player. Recall that the expected payoff e of the game is product of the matrices in the order RPC , where R is the row matrix, P is the payoff Matrix and C is the column matrix. Consider the product, e = R P C = [ x y z t ] [ 0 − 1 5 2 − 2 4 0 3 0 1 0 − 5 ] [ 1 3 1 3 1 3 ] Recall that for matrices A with dimension m × n and B with dimension k × l , the product AB is defined if the number of columns in A equal number of columns in B , n = k . For matrix R , Since, number of rows in [ x y z t ] is 4 and number of columns is 1, Substitute 4 for m and 1 for n in m × n . Thus, dimension of [ x y z t ] is 1 × 4 . For matrix P , Since, number of rows in [ 0 − 1 5 2 − 2 4 0 3 0 1 0 − 5 ] is 4 and number of columns is 3 . Substitute 4 for m and 3 for n in m × n . Thus, dimension of [ 0 − 1 5 2 − 2 4 0 3 0 1 0 − 5 ] is 4 × 3 . For matrix C , Since, number of rows in [ 1 3 1 3 1 3 ] is 3 and number of columns is 1, Substitute 3 for m and 1 for n in m × n . Thus, dimension of [ 1 3 1 3 1 3 ] is 3 × 1 . Since, number of columns in R is equal to the number of rows in P , the product RP is defined. Recall that for matrices A with dimension m × n and B with dimension n × k , the product AB is the matrix of dimension m × k and i j t h entry of AB is the sum of product of corresponding entries of row i of A and column j of B . Since dimension of first matrix is 4 × 3 and dimension of second matrix is 3 × 1 , the dimension of the product is 4 × 1 . Simplify the product [ 0 − 1 5 2 − 2 4 0 3 0 1 0 − 5 ] [ 1 3 1 3 1 3 ]

Finite Mathematics and Applied Calculus (MindTap Course List)

7th Edition

ISBN: 9781337274203

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 5.4, Problem 6E

To determine

To calculate: The optimal pure strategy for the other player of the game with the payoff matrix P=[0−152−2403010−5] and mixed strategies supplied where C=[131313]T.

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Chapter 5.4, Problem 5E

Chapter 5.4, Problem 7E

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Finite Mathematics and Applied Calculus (MindTap Course List)

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The optimal pure strategy for the other player of the game with the payoff matrix P = [ 0 − 1 5 2 − 2 4 0 3 0 1 0 − 5 ] and mixed strategies supplied where C = [ 1 3 1 3 1 3 ] T .

Solution Summary: The author calculates the optimal pure strategy for the other player of the game with the payoff matrix P=left[cc

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To calculate: The optimal pure strategy for the other player of the game with the payoff matrix P=[0−152−2403010−5] and mixed strategies supplied where C=[131313]T.

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Chapter 5 Solutions