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Engineering Computer Science Operations Research : Applications and Algorithms

Verification: Consider the following matrix, A = [ 1 2 3 4 5 6 7 8 9 ] The second-row cofactors are given below: A 21 = [ 2 3 8 9 ] Therefore, we get det ( A 21 ) = 18 − 24 = − 6 Now, we have A 22 = [ 1 3 7 9 ] Therefore, det ( A 22 ) = 9 − 21 = − 12 We have A 23 = [ 1 2 7 8 ] Therefore, det ( A 23 ) = 8 − 14 = − 6 Therefore, the expansion is , det ( A ) = ( − 1 ) 2 + 1 a 21 det ( A 21 ) + ( − 1 ) 2 + 2 a 22 det ( A 22 ) + ( − 1 ) 2 + 3 a 23 det ( A 23 ) Substituting the values, we get det ( A ) = − ( 4 ) ( − 6 ) + 5 ( − 12 ) − 6 ( − 6 )

Verification: Consider the following matrix, A = [ 1 2 3 4 5 6 7 8 9 ] The second-row cofactors are given below: A 21 = [ 2 3 8 9 ] Therefore, we get det ( A 21 ) = 18 − 24 = − 6 Now, we have A 22 = [ 1 3 7 9 ] Therefore, det ( A 22 ) = 9 − 21 = − 12 We have A 23 = [ 1 2 7 8 ] Therefore, det ( A 23 ) = 8 − 14 = − 6 Therefore, the expansion is , det ( A ) = ( − 1 ) 2 + 1 a 21 det ( A 21 ) + ( − 1 ) 2 + 2 a 22 det ( A 22 ) + ( − 1 ) 2 + 3 a 23 det ( A 23 ) Substituting the values, we get det ( A ) = − ( 4 ) ( − 6 ) + 5 ( − 12 ) − 6 ( − 6 )

Operations Research : Applications and Algorithms

Operations Research : Applications and Algorithms

4th Edition

ISBN: 9780534380588

Author: Wayne L. Winston

Publisher: Brooks Cole

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Expert Solution & Answer

Chapter 2.6, Problem 1P

Explanation of Solution

Verification:

Consider the following matrix,

A=[1    2    34    5    67    8    9]

The second-row cofactors are given below:

A21=[2    38    9]

Therefore, we get

det(A21)=18−24             =−6

Now, we have

A22=[1    37    9]

Therefore,

det(A22)=9−21             =−12

We have

A23=[1    27    8]

Therefore,

det(A23)=8−14             =−6

Therefore, the expansion is ,

det(A)=(−1)2+1a21det(A21)+(−1)2+2a22det(A22)+(−1)2+3a23det(A23)

Substituting the values, we get

det(A)=−(4)(−6)+5(−12)−6(−6)

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Chapter 2.5, Problem 11P

Chapter 2.6, Problem 2P

Chapter 2 Solutions

Operations Research : Applications and Algorithms

Ch. 2.1 - Prob. 1P Ch. 2.1 - Prob. 2P Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - Prob. 7P Ch. 2.2 - Prob. 1P Ch. 2.3 - Prob. 1P Ch. 2.3 - Prob. 2P

Ch. 2.3 - Prob. 3P Ch. 2.3 - Prob. 4P Ch. 2.3 - Prob. 5P Ch. 2.3 - Prob. 6P Ch. 2.3 - Prob. 7P Ch. 2.3 - Prob. 8P Ch. 2.3 - Prob. 9P Ch. 2.4 - Prob. 1P Ch. 2.4 - Prob. 2P Ch. 2.4 - Prob. 3P Ch. 2.4 - Prob. 4P Ch. 2.4 - Prob. 5P Ch. 2.4 - Prob. 6P Ch. 2.4 - Prob. 7P Ch. 2.4 - Prob. 8P Ch. 2.4 - Prob. 9P Ch. 2.5 - Prob. 1P Ch. 2.5 - Prob. 2P Ch. 2.5 - Prob. 3P Ch. 2.5 - Prob. 4P Ch. 2.5 - Prob. 5P Ch. 2.5 - Prob. 6P Ch. 2.5 - Prob. 7P Ch. 2.5 - Prob. 8P Ch. 2.5 - Prob. 9P Ch. 2.5 - Prob. 10P Ch. 2.5 - Prob. 11P Ch. 2.6 - Prob. 1P Ch. 2.6 - Prob. 2P Ch. 2.6 - Prob. 3P Ch. 2.6 - Prob. 4P Ch. 2 - Prob. 1RP Ch. 2 - Prob. 2RP Ch. 2 - Prob. 3RP Ch. 2 - Prob. 4RP Ch. 2 - Prob. 5RP Ch. 2 - Prob. 6RP Ch. 2 - Prob. 7RP Ch. 2 - Prob. 8RP Ch. 2 - Prob. 9RP Ch. 2 - Prob. 10RP Ch. 2 - Prob. 11RP Ch. 2 - Prob. 12RP Ch. 2 - Prob. 13RP Ch. 2 - Prob. 14RP Ch. 2 - Prob. 15RP Ch. 2 - Prob. 16RP Ch. 2 - Prob. 17RP Ch. 2 - Prob. 18RP Ch. 2 - Prob. 19RP Ch. 2 - Prob. 20RP Ch. 2 - Prob. 21RP Ch. 2 - Prob. 22RP