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(c) a1;11 denotes the element at the intersection of the rst row and the eleventh column, whereas a11;1 denotes the element at the intersection of the eleventh row and the rst column. Note: The answers to problems 25 through 29 all follow from eq. (36). 25. (a) 12 8 4, answer. (b) 3 2 8 30, answer. 26. 24 6 3 2 4 6 27, answer. 27. 20 60 40= 8 5, answer. 20 28
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28. 10x2 y y2 y 10x2 y , answer. 29. x 2 2 x 4 x 5 x2 4x 24, answer. Note: The solutions to problems 30 through 34 follow the 3-step procedure of section 3.3. 30. (1) Here, a11 3, a21 5, a31 1: 2 a11 A11 a11 1 M11 a21 A21 a21 1 3 M21 a31 A31 a31 1 4 M31
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   7 4    a11 M11 3  3 21 8 39  2 3    6 1   a21 M21 5  5 18 2 100  2 3    6 1   a31 M31   24 7 31  7 4 
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(3) D 39 100 31 108, answer. 31. (1) Here a31 1, a32 2, a33 3 2 a31 A31 a31 1 M31 a32 A32 a32 1 5 M32 a33 A33 a33 1 6 M33
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 1  a31 M31  24 7 31 4     3 1   a32 M32 2  2 12 5 14  5 4     3 6   a33 M33 3  3 21 30 153  5 7   6   7
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D 31 14 153 108, answer.
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32. The easiest way is to expand the determinant in terms of the elements of the third column, because two of the three elements there are zeros. Thus, since a13 0,
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a23 0, and a33 4, we have D 0 0 a33 1 M33
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 6 4 2  1  88; answer: 4
33. (1) Let us use the second column (because it contains two zeros). Thus we have a12 2, a22 0, a32 1, a42 0: (2)* Since a22 0 and also a42 0, we have      1 2 2  2 5 4          a12 M12 2 3 3 0  and a32 M32 1  1 2 2      1 6 3   1 6 3 Let D1 and D2 be the values of the two determinants above, and let us expand both in terms of the elements of the third column. Doing this gives us !     1 2  3 3      D1 2 2  2 30 27 6  3  3 3   1 6        2 5  2 5  1 2       D2 4  4 8 2 17 3 1 69  3  2 1 2  1 6   1 6  (3) D D1 D2 6 69 63, answer. 34. Let us begin by expanding the given determinant in terms of the elements of the fourth row. Since all the elements of the fourth row except the 5 are zeros, the value of the determinant reduces to the value of the single third-order determinant   1 3 2     D 5 6 1 3     4 2 0 Now expand the above result in terms of, let us say, the third column; thus   6 D 5 2 4    1  3  1 4  2  3  5 32 42 370; answer: 2 
35. First factor 6 from column 1 and 5 from column 3. Then factor 4 from the second row. We can then, if we wish, expand the determinant in terms of the minors of the rst column; thus    1 3 1  !         3 1   4 6   3 1          1 4 6  120  6 5 4    2      4 6   1 5  1 5    2 1 5 120 26 16 28 2160; answer:
* Step (2) can also be stated as follows. Let a denote any element in a determinant, and let aM denote the product of a and its minor determinant M. Let s denote the sum of the number of the row and the number of the column in which a is located; if s is an even number write aM, but if s is an odd number write aM.
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