quantitative analysis

# A evaluating a 2 2 determinant 12 11 2 a a a a the

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(a) EVALUATING a 2 2 Determinant Let 22 21 12 11 2 a a a a The value of this determinant is defined as: 21 12 22 11 22 21 12 11 a a a a a a a a For example, 7 12 5 ) 6 )( 2 ( ) 1 )( 5 ( 1 2 6 5

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COSTING AND QUANTITATIVE TECHNIQUES 360 (b) Expansion of A 3 3 DETERMINANT Consider the following 3 3 determinant. 33 32 31 23 22 21 13 12 11 a a a a a a a a a ----------------------------- 16.4.1 This can be expanded by reducing it into a linear combination of 2 2 determinants using a row or a column of the determinant. The following 2 2 determinants are commonly used: rs A = determinant obtained from by deleting from A its r th row and s th column. For example, 33 32 23 22 11 A a a a a , 33 31 23 21 12 A a a a a , 32 31 22 21 13 A a a a a , 33 32 13 12 21 A a a a a 33 31 13 11 22 A a a a a , 32 31 12 11 23 A a a a a , 23 22 23 12 31 A a a a a , 23 21 13 11 32 A a a a a 22 21 12 11 33 A a a a a These can be put together as a matrix of cofactors: i.e Matrix of cofactors = 33 32 31 23 22 21 13 12 11 A A A A A A A A A For example, to expand the determinant using (16.5.1) and the 1 st row, we have = 13 13 3 1 12 12 2 1 11 11 1 1 ) 1 ( ) 1 ( ) 1 ( A a A a A a To expand by the second column, we get, = 32 32 3 2 22 22 2 2 12 12 1 2 ) 1 ( ) 1 ( ) 1 ( A a A a A a
INTRODUCTION TO MATRICES 361 ILLUSTRATION 16-1 Evaluate 6 0 4 2 0 8 1 5 3 SUGGESTED SOLUTION 16-1 To expand by the first row, we need: 0 6 0 2 0 A 11 , 40 6 4 2 8 A 12 , 0 0 4 0 8 A 13 where, for example, A 12 is the determinant obtained by deleting the 1 st row and the 2 nd column of . Hence, A 11 = 0, A 12 = 40, A 13 = 0 Expansion by the first row gives: 13 13 3 1 12 12 2 1 11 11 1 1 ) 1 ( ) 1 ( ) 1 ( A a A a A a ………. 16.4.2 Substituting for 13 13 12 12 11 11 A , , A , , A , a a a , we get, = (1) (3) (0) + (-1) (5) (40) + (1) (1) (0) = 0 - 200 + 0 = -200 Expanding by the second column is the easiest since two of the entries in column 2 are zero. Expanding by column 2 gives = 32 32 3 3 22 22 2 2 12 12 2 1 ) 1 ( ) 1 ( ) 1 ( A a A a A a …….. 16.4.3 But 0 32 22 a a Hence , from 16.4.3, we have, 12 12 A a …………………. 16.4.4 where 12 a = 5 and A 12 = 40 8 48 6 4 2 8 Substituting these in 16.4.3 12 12 A a = (-1) (5) (40) = -200

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COSTING AND QUANTITATIVE TECHNIQUES 362 Again, expand by the 3 rd row = 33 33 3 3 32 32 2 3 31 31 1 3 ) 1 ( ) 1 ( ) 1 ( A a A a A a ….. 16.4.5 where 31 a = 4, 32 a = 0, 33 a = 6 and A 31 = 2 0 1 5 = 10, A 32 = 6 4 1 3 = 14, A 33 = 0 8 5 3 = - 40 So, the expansion of by the 2 nd column gives = 33 33 32 32 31 31 ) 1 ( A a A a A a …………. 16.4.6 = (4) (10) + (0) (14) + (6) (-40) = 40 240 = - 200 as before. ILLUSTRATION 16-2 Evaluate 1 2 7 6 5 1 3 1 4 SUGGESTED SOLUTION 16-2 Let us expand by the second column. = 32 3 3 22 2 2 12 2 1 ) 2 ( ) 1 ( ) 5 ( ) 1 ( ) 1 ( ) 1 ( A A A ……. 16.4.7 where A 12 = 4 42 1 1 7 6 1 ; A 22 = 25 21 4 1 2 3 4 ; and A 12 = 27 3 24 6 1 3 4 we get, = 32 32 3 3 22 22 2 2 12 12 2 1 ) 1 ( ) 1 ( ) 1 ( A a A a A a
INTRODUCTION TO MATRICES 363 = (-1) (1) (-41) + (1) (5) (25) + (-1) (-2) (27) ………… 16.4.8 = 41 + 125 + 54 = 220 Next, expand by the 3 rd row to get = 33 33 3 3 32 32 2 3 31 31 1 3 ) 1 ( ) 1 ( ) 1 ( A a A a A a ………….. 16.4.9 A 31 = 6 5 3 1 = 27; A 32 = 6 1 3 4 = 27; A 33 = 5 1 1 4 = 19 and 31 a = 7, 32 a = - 2, 33 a = 1. = (1) (7) (21) + (-1) (-2) (27) + (1) (1) (19) = 147 + 54 + 19 = 220 as before Conclusion No matter what row or column is used for the expansion, the end result is the same and it is the value of the determinant. In practice, any row or any column can be used for the expansion.

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